Glimpses of Algebra and Geometry, Second Edition (Undergraduate

Undergraduate Texts in Mathematics Readings in Mathematics

Editors S. Axler F.W. Gehring K.A. Ribet Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

This page intentionally left blank Gabor Toth

Glimpses of Algebra and Geometry

Second Edition

With 183 Illustrations, Including 18 in Full Color Gabor Toth Department of Mathematical Sciences Rutgers University Camden, NJ 08102 USA [email protected] Editorial Board S. Axler F.W. Gehring K.A. Ribet Mathematics Department Mathematics Department Mathematics Department San Francisco State East Hall University of California, University University of Michigan Berkeley San Francisco, CA 94132 Ann Arbor, MI 48109 Berkeley, CA 94720-3840 USA USA USA

Front cover illustration: The regular compound of five tetrahedra given by the face- planes of a colored icosahedron. The circumscribed dodecahedron is also shown. Computer graphic made by the author using Geomview. Back cover illustration: The regular compound of five cubes inscribed in a dodecahedron. Computer graphic made by the author using Mathematica.

Mathematics Subject Classification (2000): 15-01, 11-01, 51-01

Library of Congress Cataloging-in-Publication Data Toth, Gabor, Ph.D. Glimpses of algebra and geometry/Gabor Toth.—2nd ed. p. cm. — (Undergraduate texts in mathematics. Readings in mathematics.) Includes bibliographical references and index. ISBN 0-387-95345-0 (hardcover: alk. pape 1. Algebra. 2. Geometry. I. Title. II. Series. QA154.3 .T68 2002 512′.12—dc21 2001049269 Printed on acid-free paper.

 2002, 1998 Springer-Verlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information stor- age and retrieval, electronic adaptation, computer software, or by similar or dissimi- lar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Production managed by Francine McNeill; manufacturing supervised by Jeffrey Taub. Typeset from the author’s 2e files using Springer’s UTM style macro by The Bartlett Press, Inc., Marietta, GA. Printed and bound by Hamilton Printing Co., Rensselaer, NY. Printed in the United States of America. 987654321 ISBN 0-387-95345-0 SPIN 10848701 Springer-Verlag New York Berlin Heidelberg A member of BertelsmannSpringer Science+Business Media GmbH This book is dedicated to my students. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

This page intentionally left blank Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Preface to the . . . . . Second Edition . . .

Since the publication of the Glimpses in 1998, I spent a considerable amount of time collecting “mathematical pearls” suitable to add to the original text. As my collection grew, it became clear that a major revision in a second edition needed to be considered. In addition, many readers of the Glimpses suggested changes, clarifi- cations, and, above all, more examples and worked-out problems. This second edition, made possible by the ever-patient staff of Springer-Verlag New York, Inc., is the result of these efforts. Al- though the general plan of the book is unchanged, the abundance of topics rich in subtle connections between algebra and geometry compelled me to extend the text of the first edition considerably. Throughout the revision, I tried to do my best to avoid the inclusion of topics that involve very difficult ideas. The major changes in the second edition are as follows: 1. An in-depth treatment of root formulas solving quadratic, cubic, and quartic equations a` la van der Waerden has been given in a new section. This can be read independently or as preparation for the more advanced new material encountered toward the later parts of the text. In addition to the Bridge card symbols, the dagger † has been introduced to indicate more technical material than the average text. vii Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

viii Preface to the Second Edition

2. As a natural continuation of the section on the Platonic solids, a detailed and complete classification of finite Mobius¨ groups ala` Klein has been given with the necessary background material, such as Cayley’s theorem and the Riemann–Hurwitz relation. 3. One of the most spectacular developments in algebra and geometry during the late nineteenth century was Felix Klein’s theory of the icosahedron and his solution of the irreducible quintic in terms of hypergeometric functions. A quick, direct, and modern approach of Klein’s main result, the so-called Normalformsatz, has been given in a single large section. This treatment is independent of the material in the rest of the book, and is suitable for enrichment and undergraduate/graduate research projects. All known approaches to the solution of the irreducible quintic are technical; I have chosen a geometric approach based on the construction of canonical quintic resolvents of the equation of the icosahedron, since it meshes well with the treatment of the Platonic solids given in the earlier part of the text. An algebraic approach based on the reduction of the equation of the icosahedron to the Brioschi quintic by Tschirnhaus transformations is well documented in other textbooks. Another section on polynomial invariants of finite Mobius¨ groups, and two new appendices, containing preparatory material on the hypergeometric differential equation and Galois theory, facilitate the understanding of this advanced material. 4. The text has been upgraded in many places; for example, there is more material on the congruent number problem, the stereographic projection, the Weierstrass ℘-function, projective spaces, and isometries in space. 5. The new Web site at http://mathsgi01.rutgers.edu/∼gtoth/ Glimpses/ containing various text files (in PostScript and HTML formats) and over 70 pictures in full color (in gif format) has been created. 6. The historical background at many places of the text has been made more detailed (such as the ancient Greek approximations of π), and the historical references have been made more precise. 7. An extended solutions manual has been created containing the solutions of 100 problems. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Preface to the Second Edition ix

I would like to thank the many readers who suggested improve- ments to the text of the first edition. These changes have all been incorporated into this second edition. I am especially indebted to Hillel Gauchman and Martin Karel, good friends and colleagues, who suggested many worthwhile changes. I would also like to express my gratitude to Yukihiro Kanie for his careful reading of the text and for his excellent translation of the first edition of the Glimpses into Japanese, published in early 2000 by Springer- Verlag, Tokyo. I am also indebted to April De Vera, who upgraded the list of Web sites in the first edition. Finally, I would like to thank Ina Lindemann, Executive Editor, Mathematics, at Springer-Verlag New York, Inc., for her enthusiasm and encouragement throughout the entire project, and for her support for this early second edition.

Camden, New Jersey Gabor Toth Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

This page intentionally left blank Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Preface to the . . . . . First Edition . . .

Glimpse: 1. a very brief passing look, sight or view. 2. a momentary or slight appearance. 3. a vague idea or inkling. —Random House College Dictionary

At the beginning of fall 1995, during a conversation with my re- spected friend and colleague Howard Jacobowitz in the Octagon Dining Room (Rutgers University, Camden Campus), the idea emerged of a “bridge course” that would facilitate the transition between undergraduate and graduate studies. It was clear that a course like this could not concentrate on a single topic, but should browse through a number of mathematical disciplines. The selection of topics for the Glimpses thus proved to be of utmost importance. At this level, the most prominent interplay is manifested in some easily explainable, but eventually subtle, connections between number theory, classical geometries, and modern algebra. The rich, fascinating, and sometimes puzzling interactions of these mathematical disciplines are seldom contained in a medium-size undergraduate textbook. The Glimpses that follow make a humble effort to ﬁll this gap.

xi Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

xii Preface to the First Edition

The connections among the disciplines occur at various levels in the text. They are sometimes the main topics, such as Rational- ity and Elliptic Curves (Section 3), and are sometimes hidden in problems, such as the spherical geometric proof of diagonalization of Euclidean isometries (Problems 1 to 2, Section 16), or the proof of Euler’s theorem on convex polyhedra using linear algebra (Prob- lem 9, Section 20). Despite numerous opportunities throughout the text, the experienced reader will no doubt notice that analysis had to be left out or reduced to a minimum. In fact, a major source of difficulties in the intense 8-week period during which I produced the first version of the text was the continuous cutting down of the size of sections and the shortening of arguments. Further- more, when one is comparing geometric and algebraic proofs, the geometric argument, though often more lengthy, is almost always more revealing and thereby preferable. To strive for some originality, I occasionally supplied proofs out of the ordinary, even at the “expense” of going into calculus a bit. To me, “bridge course” also meant trying to shed light on some of the links between the first recorded intellectual attempts to solve ancient problems of number theory, geometry, and twentieth-century mathematics. Ignoring detours and sidetracks, the careful reader will see the continuity of the lines of arguments, some of which have a time span of 3000 years. In keeping this continuity, I eventually decided not to break up the Glimpses into chapters as one usually does with a text of this size. The text is, nevertheless, broken up into subtexts corresponding to various levels of knowledge the reader possesses. I have chosen the card symbols ♣, ♦, ♥, ♠ of Bridge to indicate four levels that roughly correspond to the following: ♣ College Algebra; ♦ Calculus, Linear Algebra; ♥ Number Theory, Modern Algebra (elementary level), Geometry; ♠ Modern Algebra (advanced level), Topology, Complex Variables. Although much of ♥ and ♠ can be skipped at first reading, I encour- age the reader to challenge him/herself to venture occasionally into these territories. The book is intended for (1) students (♣ and ♦) who wish to learn that mathematics is more than a set of tools (the way sometimes calculus is taught), (2) students (♥ and ♠) who Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Preface to the First Edition xiii love mathematics, and (3) high-school teachers (⊂{♣, ♦, ♥, ♠}) who always had keen interest in mathematics but seldom time to pursue the technicalities. Reading what I have written so far, I realize that I have to make one point clear: Skipping and reducing the size of subtle arguments have the inherent danger of putting more weight on intuition at the expense of precision. I have spent a considerable amount of time polishing intuitive arguments to the extent that the more experienced reader can make them withstand the ultimate test of mathematical rigor. Speaking (or rather writing) of danger, another haunted me for the duration of writing the text. One of my favorite authors, Iris Murdoch, writes about this in The Book and the Brotherhood,in which Gerard Hernshaw is badgered by his formidable scholar Lev- quist about whether he wanted to write mediocre books out of great ones for the rest of his life. (Tolearn what Gerard’s answer was, you need to read the novel.) Indeed, a number of textbooks iuenced me when writing the text. Here is a sample: 1. M. Artin, Algebra, Prentice-Hall, 1991; 2. A. Beardon, The Geometry of Discrete Groups, Springer-Verlag, 1983; 3. M. Berger, Geometry I–II, Springer-Verlag, 1980; 4. H.S.M. Coxeter, Introduction to Geometry, Wiley, 1969; 5. H.S.M. Coxeter, Regular Polytopes, Pitman, 1947; 6. D. Hilbert and S. Cohn-Vossen, Geometry and Imagination, Chelsea, 1952. 7. J. Milnor, Topology from the Differentiable Viewpoint, The Univer- sity Press of Virginia, 1990; 8. I. Niven, H. Zuckerman, and H. Montgomery, An Introduction to the Theory of Numbers, Wiley, 1991; 9. J. Silverman and J. Tate, Rational Points on Elliptic Curves, Springer-Verlag, 1992. Although I (unavoidably) use a number of by now classical arguments from these, originality was one of my primary aims. This book was never intended for comparison; my hope is that the Glimpses may trigger enough motivation to tackle these more advanced textbooks. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

xiv Preface to the First Edition

Despite the intertwining nature of the text, the Glimpses contain enough material for a variety of courses. For example, a shorter version can be created by taking Sections 1 to 10 and Sections 17 and 19 to 23, with additional material from Sections 15 to 16 (treating Fuchsian groups and Riemann surfaces marginally via the examples) when needed. A nonaxiomatic treatment of an undergraduate course on geometry is contained in Sections 5 to 7, Sections 9 to 13, and Section 17. The Glimpses contain a lot of computer graphics. The material can be taught in the traditional way using slides, or interactively in a computer lab or teaching facility equipped with a PC or a workstation connected to an LCD-panel. Alternatively, one can create a graphic library for the illustrations and make it accessi- ble to the students. Since I have no preference for any software packages (although some of them are better than others for particular purposes), I used both Maple®1 and Mathematica®2 to create the illustrations. In a classroom setting, the link of either of these to Geomview3 is especially useful, since it allows one to manipu- late three-dimensional graphic objects. Section 17 is highly graphic, and I recommend showing the students a variety of slides or three- dimensional computer-generated images. Animated graphics can also be used, in particular, for the action of the stereographic projection in Section 7, for the symmetry group of the pyramid and the prism in Section 17, and for the cutting-and-pasting technique in Sections 16 and 19. These Maple® text ﬁles are downloadable from my Web sites

http://carp.rutgers.edu/math-undergrad/science-vision.html and http://mathsgi01.rutgers.edu/∼gtoth/.

Alternatively, to obtain a copy, write an e-mail message to

[email protected]

1Maple is a registered trademark of Waterloo Maple, Inc. 2Mathematica is a registered trademark of Wolfram Research, Inc. 3A software package downloadable from the Web site: http://www.geom.umn.edu. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Preface to the First Edition xv or send a formatted disk to Gabor Toth, Department of Mathemat- ical Sciences, Rutgers University, Camden, NJ 08102, USA.

A great deal of information, interactive graphics, animations, etc., are available on the World Wide Web. I highly recommend scheduling at least one visit to a computer or workstation lab and explaining to the students how to use the Web. In fact, at the ﬁrst implementation of the Glimpses at Rutgers, I noticed that my students started spending more and more time at various Web sites related to the text. For this reason, I have included a list of recom- mended Web sites and ﬁlms at the end of some sections. Although hundreds of Web sites are created, upgraded, and terminated daily, every effort has been made to list the latest Web sites currently available through the Interent.

Camden, New Jersey Gabor Toth Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

This page intentionally left blank Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Acknowledgments . . . . .

The second half of Section 20 on the four color theorem was written by Joseph Gerver, a colleague at Rutgers. I am greatly indebted to him for his contribution and for sharing his insight into graph theory. The first trial run of the Glimpses at Rutgers was during the first six weeks of summer 1996, with an equal number of undergraduate and graduate students in the audience. In fall 1996, I also taught undergraduate geometry from the Glimpses, covering Sec- tions 1 to 10 and Sections 17 and 19 to 23. As a result of the students’ dedicated work, the original manuscript has been revised and cor- rected, some of the arguments have been polished, and some extra topics have been added. It is my pleasure to thank all of them for their participation, enthusiasm, and hard work. I am particularly indebted to Jack Fistori, a mathematics education senior at Rutgers, who carefully revised the final version of the manuscript, making numerous worthwhile changes. I am also indebted to Susan Carter, a graduate student at Rutgers, who spent innumerable hours at the workstation to locate suitable Web sites related to the Glimpses. In summer 1996, I visited the Geometry Center at the University of Minnesota. I lectured about the Glimpses to an audience consisting of undergraduate and graduate students and high-school teachers. I wish to thank them for their valuable comments, which I took xvii Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

xviii Acknowledgments

into account in the final version of the manuscript. I am especially indebted to Harvey Keynes, Education Director of the Geometry Center, for his enthusiastic support of the Glimpses. During my stay, I produced a 10-minute film Glimpses of the Five Platonic Solids with Stuart Levy, whose dedication to the project surpassed all my expectations. The typesetting of the manuscript started when I gave the first 20 pages to Betty Zubert as material with which a to practice LTEX. As the manuscript grew beyond any reasonable size, it is my pleasure to record my thanks to her for providing in- exhaustible energy that turned 300 pages of chicken scratch into a fine document.

Camden, New Jersey Gabor Toth Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Contents . . . . .

Preface to the Second Edition vii Preface to the First Edition xi Acknowledgments xvii Section 1 “A Number Is a Multitude Composed of Units”—Euclid 1 Problems 6 Web Sites 6 Section 2 “... There Are No Irrational Numbers at All”—Kronecker 7 Problems 21 Web Sites 25 Section 3 Rationality, Elliptic Curves, and Fermat’s Last Theorem 26 Problems 52 Web Sites 54 Section 4 Algebraic or Transcendental? 55 Problems 60 xix Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

xx Contents

Section 5 Complex Arithmetic 62 Problems 71 Section 6 Quadratic, Cubic, and Quantic Equations 72 Problems 80 Section 7 Stereographic Projection 83 Problems 88 Web Site 89 Section 8 Proof of the Fundamental Theorem of Algebra 90 Problems 93 Web Site 95 Section 9 Symmetries of Regular Polygons 96 Problems 105 Web Sites 106 Section 10 Discrete Subgroups of Iso (R2)107 Problems 120 Web Sites 121 Section 11 Mobius¨ Geometry 122 Problems 130 Section 12 Complex Linear Fractional Transformations 131 Problems 137 Section 13 “Out of Nothing I Have Created a New Universe”—Bolyai 139 Problems 156 Section 14 Fuchsian Groups 158 Problems 171 Section 15 Riemann Surfaces 173 Problems 197 Web Site 198 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Contents xxi

Section 16 General Surfaces 199 Problems 208 Web Site 208 Section 17 The Five Platonic Solids 209 Problems 248 Web Sites 254 Film 254 Section 18 Finite Mobius¨ Groups 255

Section 19 Detour in Topology: Euler–Poincare´ Characteristic 266 Problems 278 Film 278 Section 20 Detour in Graph Theory: Euler, Hamilton, and the Four Color Theorem 279 Problems 294 Web Sites 297 Section 21 Dimension Leap 298 Problems 304 Section 22 Quaternions 305 Problems 315 Web Sites 316 Section 23 Back to R3!317 Problems 328 Section 24 Invariants 329 Problem 344 Section 25 The Icosahedron and the Unsolvable Quintic 345 A. Polyhedral Equations 346 B. Hypergeometric Functions 348 C. The Tschirnhaus Transformation 351 D. Quintic Resolvents of the Icosahedral Equation 355 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

xxii Contents

E. Solvability of the Quintic a` la Klein 363 F. Geometry of the Canonical Equation: General Considerations 365 G. Geometry of the Canonical Equation: Explicit Formulas 369 Problems 377 Section 26 The Fourth Dimension 380 Problems 394 Film 395 Appendix A Sets 397 Appendix B Groups 399 Appendix C Topology 403 Appendix D Smooth Maps 407 Appendix E The Hypergeometric Differential Equation and the Schwarzian 409 Appendix F Galois Theory 419 Solutions for 100 Selected Problems 425 Index 443 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

. . . . . “A Number Is a . . . . . Multitude Composed . . . 1 . SECTION . of Units”—Euclid

♣ We adopt Kronecker’s phrase: “God created the natural numbers, and all the rest is the work of man,” and start with the set N {1, 2, 3, 4, 5, 6,...} of all natural numbers. Since the sum of two natural numbers is again a natural number, N carries the operation1 of addition + : N × N → N.

Remark. Depicting natural numbers by arabic numerals is purely traditional. Romans might prefer N {I, II, III, IV, V, VI,...}, and computers work with N {1, 10, 11, 100, 101, 110,...}. Notice that converting a notation into another is nothing but an isomorphism between the respective systems. Isomorphism respects

1If needed, please review “Sets” and “Groups” in Appendices A and B. 1 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

2 1. “A Number Is a Multitude Composed of Units”—Euclid

addition; for example, 29+33 62 is the same as XXIX +XXXIII LXII or 11101 + 100001 111110.

From the point of view of group theory, N is a failure; it does not have an identity element (that we would like to call zero) and no element has an inverse. We remedy this by extending N to the (additive) group of integers Z {0, ±1, ±2, ±3, ±4, ±5, ±6,...}. Z also carries the operation of multiplication × : Z × Z → Z. Since distributivity holds, Z forms a ring with respect to addition and multiplication. Although we have 1 as the identity element with respect to ×, we have no hope for Z to be a multiplicative group; remember the saying: “Thou shalt not divide by zero!” To remedy this, we delete the ominous zero and consider Z# Z −{0} {±1, ±2, ±3, ±4, ±5, ±6,...}. The requirement that integers have inverses gives rise to fractions or, more appropriately, rational numbers: Q Q # ∪{0} {a/b | a, b ∈ Z#}∪{0}, where we put the zero back to save the additive group structure. All that we learned in dealing with fractions can be rephrased elegantly by saying that Q is a ﬁeld: Q is an additive group, Q # is an abelian (i.e., commutative) multiplicative group, and addition and multiplication are connected through distributivity. After having created Z and Q , the direction we take depends largely on what we wish to study. In elementary number theory, when studying divisibility properties of integers, we consider, for a given n ∈ N, the (additive) group Zn of integers modulo n. The simplest way to understand

Zn Z/nZ {[0], [1],...,[n − 1]} is to start with Z and to identify two integers a and b if they differ by a multiple of n. This identification is indicated by the square bracket; [a] means a plus all multiples of n. Clearly, no numbers are identified among 0, 1,...,n− 1, and any integer is identified Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

1. “A Number Is a Multitude Composed of Units”—Euclid 3

0 n 2n Figure 1.1

with exactly one of these. The (additive) group structure is given by the usual addition in Z. More explicitly, [a] + [b] [a + b], a, b ∈ Z. Clearly, [0] is the zero element in Zn, and −[a] [−a] is the additive inverse of [a] ∈ Zn. Arithmetically, we use the division algorithm to ﬁnd the quotient q and the remainder 0 ≤ r

a qn + r, and set [a] [r]. The geometry behind this equality is clear. Con- sider the multiples of n, nZ ⊂ Z,asaone-dimensional lattice (i.e., an infinite string of equidistantly spaced points) in R as in Figure 1.1. Now locate a and its closest left neighbor qn in nZ (Figure 1.2). The distance between qn and a is r, the latter between 0 and n − 1. Since a and r are to be identified, the following geometric picture emerges for Zn:WrapZ around a circle infinitely many times so that the points that overlap with 0 are exactly the lattice points in nZ; this can be achieved easily by choosing the radius of the circle to be n/2π. Thus, Zn can be visualized as n equidistant points on the perimeter of a circle (Figure 1.3). Setting the center of the circle at the origin of a coordinate system on the Cartesian plane R2 such that [0] is the intersection point of the circle and the positive first axis, we see that addition in Zn corresponds to addition of angles of the corresponding vectors. A common convention is to choose the positive orientation as the way [0], [1], [2],...increase. This picture of Zn as the vertices of a regular n-sided polygon (with angular addition) will recur later on in several different contexts.

(q − 1)naqn (q + 1)n Figure 1.2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

4 1. “A Number Is a Multitude Composed of Units”—Euclid

Figure 1.3

Remark. In case you’ve ever wondered why it was so hard to learn the clock 2 in childhood, consider Z60. Why the Babylonian choice of 60? Con- sider natural numbers between 1 and 100 that have the largest possible number of small divisors.

♥ The infinite Z and its finite offsprings Zn, n ∈ N, share the basic property that they are generated by a single element, a property that we express by saying that Z and Zn are cyclic. In case of Z, this element is 1 or −1; in case of Zn, a generator is [1]. ♣ You might be wondering whether it is a good idea to reconsider multiplication in Zn induced from that of Z. The answer is yes; multiplication in Z gives rise to a well-defined multiplication in Zn by setting [a] · [b] [ab], a, b ∈ Z. Clearly, [1] is the multiplicative # identity element. Consider now multiplication restricted to Zn Zn −{[0]}. There is a serious problem here. If n is composite, that # is, n ab, a, b ∈ N, a, b ≥ 2, then [a], [b] ∈ Zn, but [a] · [b] [0]! # Thus, multiplication restricted to Zn is not even an operation.

2Actually, a number system using 60 as a base was developed by the Sumerians about 500 years before it was passed on to the Babylonians around 2000 b.c. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

1. “A Number Is a Multitude Composed of Units”—Euclid 5

We now pin our hopes on Zp, where p a prime. Elementary number theory says that if p divides ab, then p divides either a or b. # This directly translates into the fact that Zp is closed under multiplication. Encouraged by this, we now go a step further and claim # that Zp is a multiplicative group! Since associativity follows from associativity of multiplication in Z, it remains to show that each el- # ement a ∈ Zp has a multiplicative inverse. To prove this, multiply the complete list [1], [2],...,[p − 1] by [a] to obtain

[a], [2a],...,[(p − 1)a].

# By the above, these all belong to Zp. They are mutually disjoint. Indeed, assume that [ka] [la], k, l 1, 2,...,p− 1. We then have [(k − l)a] [k − l] · [a] [0], so that k l follows. Thus, the list # above gives p − 1 elements of Zp. But the latter consists of exactly p − 1 elements, so we got them all! In particular, [1] is somewhere in this list, say, [aa¯ ] [1], a¯ 1,...,p− 1. Hence, [a¯] is the multiplicative inverse of [a]. Finally, since distributivity in Zp follows from distributivity in Z, we obtain that Zp is a field for p prime. We give two applications of these ideas: one for Z3 and another 2 2 for Z4. First, we claim that if 3 divides a + b , a, b ∈ Z, then 3 divides both a and b. Since divisibility means zero remainder, all we have to count is the sum of the remainders when a2 and b2 are divided by 3. In much the same way as we divided all integers to even (2k) and odd (2k + 1) numbers (k ∈ Z), we now write a 3k, 3k + 1, 3k + 2 accordingly. Squaring, we obtain a2 9k2, 9k2 + 6k + 1, 9k2 + 12k + 4. Divided by 3, these give remainders 0 or 1, with 0 corresponding to a being a multiple of 3. The situation is the same for b2. We see that when dividing a2 + b2 by 3, the possible remainders are 0 + 0, 0 + 1, 1 + 0, 1 + 1, and the first corresponds to a and b both being multiples of 3. The first claim follows. Second, we show the important number theoretical fact that no number of the form 4m + 3 is a sum of two squares of integers. (Notice that, for m 0, this follows from the first claim or by inspection.) This time we study the remainder when a2 + b2, a, b ∈ Z, is divided by 4. Setting a 4k, 4k + 1, 4k + 2, 4k + 3, a2 gives remainders 0 or 1. As before, the possible remainders for a2 + b2 are 0 + 0, 0 + 1, 1 + 0, 1 + 1. The second claim also follows. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

6 1. “A Number Is a Multitude Composed of Units”—Euclid

♥ The obvious common generalization of the two claims above is also true and is a standard fact in number theory. It asserts that if a prime p of the form 4m +3 divides a2 +b2, a, b ∈ Z, then p divides both a and b. Aside from the obvious decomposition 2 12 + 12, the primes that are left out from our considerations are of the form 4m + 1. A deeper result in number theory states that any prime of the form 4m + 1 is always representable as a sum of squares of two integers. Fermat, in a letter to Mersenne in 1640, claimed to have a proof of this result, which was first stated by Albert Girard in 1632. The first published verification, due to Euler, appeared in 1754. We postpone the proof of this result till the end of Section 5.

Problems

1. Use the division algorithm to show that (a) the square of an integer is of the form 3a or 3a + 1, a ∈ Z; (b) the cube of an integer is of the form 7a,7a + 1 or 7a − 1, a ∈ Z. 2. Prove that if an integer is simultaneously a square and a cube, then it must be of the form 7a or 7a + 1. (Example: 82 43.)

3. (a) Show that [2] has no inverse in Z4. (b) Find all n ∈ N such that [2] has an inverse in Zn.

4. Write a ∈ N in decimal digits as a a1a2 ···an, a1,a2,...,an ∈{0, 1,...,9}, a1 0. Prove that [a] [a1 + a2 +···+an]inZ9. 5. Let p>3 be a prime, and write

1 + 1 +···+ 1 12 22 (p − 1)2 as a rational number a/b, where a, b are relatively prime. Show that p|a.

Web Sites

1. www.utm.edu/research/primes 2. daisy.uwaterloo.ca/∼alopez-o/math-faq/node10.html Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... “ There Are No . . . . . Irrational Numbers . . . 2 . SECTION . at All”—Kronecker

♣ Although frequently quoted, the epigraph to this chapter and some other statements of Kronecker on irrational numbers have been shown to be distortions.1 (See H.M. Edwards’s articles on Kronecker in History and Philosophy of Modern Mathematics (Min- neapolis, MN, 1985) 139–144, Minnesota Stud. Philos. Sci., XI, Univ. Minnesota Press, Minneapolis, MN, 1988.) In calculus, the field of rational numbers Q is insufficient for several reasons, including convergence. Thus, Q is extended to the field of real numbers R. It is visualized by passing the Cartesian Bridge2 connecting algebra and geometry; each real number (in infinite decimal representation) corresponds to a single point on the real line. Do we actually get new numbers? Here is a simple answer known to Pythagoras (c. 570–490 b.c.):

1I am indebted to Victor Pambuccian for pointing this out. 2Cartesius is the Latinized name of Rene´ Descartes, who, contrary to widespread belief, did not invent the coordinate axes, much less analytic geometry. Here we push this gossip a little further. Note that analytic geometry was born in Fermat’s Introduction to Plane and Solid Loci in 1629; although circulated from 1637 on, it was not published in Fermat’s lifetime. The notion of perpendicular coordinate axes can be traced back to Archimedes and Apollonius. Both Descartes and Fermat used coordinates but only with nonnegative values; the idea that coordinates can also take negative values is due to Newton. 7 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

8 2. “... There Are No Irrational Numbers at All”—Kronecker

Proposition√ 1. 2(the unique positive number whose square is 2) is not rational.

Proof. √ √ ¬3 Assume that 2 is rational; i.e., 2 a/b for some a, b ∈ Z.We may assume that a and b are relatively prime, since otherwise we cancel the common factors in a and b. Squaring, we get a2 2b2. A glimpse of the right-hand side shows that a2 is even. Thus a must be even, say, a 2c. Then a2 4c2 2b2. Hence b2 and b must be even. Thus 2 is a common factor of a and b. ¬

Remark. √ Replacing 2 by any prime p, we see that p is not rational. Instead of repeating the argument above, we now describe a more powerful result due to Gauss. (If you study the following proof carefully, you will see that it is a generalization of the Pythagorean argument above.) The idea√ is very simple (and will reoccur later) and is based p x2 −p on the fact that is a solution√ of the quadratic equation 0. To show irrationality of p, we study the rational solutions of this equation. More generally, assume that the polynomial equation

n P(x) c0 + c1x +···+cnx 0,cn 0,

with integer coefﬁcients c0,c1,...,cn ∈ Z, has a rational root x a/b, a, b ∈ Z. As usual, we may assume that a and b are relatively prime. Substituting, we have

n c0 + c1(a/b) +···+cn(a/b) 0.

Multiplying through by bn−1, we obtain

n−1 n−2 n−1 n c0b + c1ab +···+cn−1a + cna /b 0.

n This says that cna /b must be an integer, or equivalently, b divides n cna . Since a and b are relatively prime, we conclude that b divides

3¬ indicates indirect argument; that is, we assume that the statement is false and get (eventually) a contradiction (indicated by another ¬). Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

2. “... There Are No Irrational Numbers at All”—Kronecker 9

n cn. In a similar vein, if we multiply through by b /a,weget

n n−1 n−1 c0b /a + c1b +···+cna 0, and it follows that a divides c0. Specializing, we see that if x a/b is a solution of xn c, then b ±1 and so c (±a)n. Thus, if c is not the nth power of an integer, then xn c does not have any rational solutions. This is indeed a vast generalization of the Pythagorean argument above! Algebraically, we think of a real number as an infinite decimal. This decimal representation is unique (and thereby serves as a definition), assuming that we exclude representations terminating in a string of infinitely many 9’s; for example, instead of 1.2999 ...we write 1.3. How can we recognize the rational numbers in this representation? Writing 1/3as0.333 ...gives a clue to the following:

Proposition 2. An inﬁnite decimal represents a rational number iff it terminates or repeats.

Proof. We ﬁrst need to evaluate the ﬁnite geometric sum

1 + x + x2 +···+xn.

For x 1, this is n + 1, hence we may assume that x 1. This sum is telescopic; in fact, after multiplying through by 1 − x, everything cancels except the ﬁrst and last term. We obtain

(1 − x)(1 + x + x2 +···+xn) 1 − xn+1, or equivalently

1 − xn+1 1 + x + x2 +···+xn ,x 1. 1 − x Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

10 2. “... There Are No Irrational Numbers at All”—Kronecker

Letting n →∞and assuming that |x| < 1 to assure convergence, we arrive at the geometric series formula4 1 1 + x + x2 ··· , |x| < 1. 1 − x We now turn to the proof. Since terminating decimals are clearly rationals, and after multiplying through by a power of 10 if necessary, we may assume that our repeating decimal representation looks like

0.a1a2 ···aka1a2 ···aka1a2 ···ak ···,

where the decimal digits ai are between 0 and 9. We rewrite this as −k −2k −3k a1a2 ···ak(10 + 10 + 10 +···)

−k −k −k 2 a1 ···ak10 (1 + 10 + (10 ) +···) a ···a −k a ···a 1 k10 1 k 1 − 10−k 10k − 1 where we used the geometric series formula. The number we arrive at is clearly rational, and we are done. The converse statement follows from the division algorithm. Indeed, if a, b are integers and a is divided by b, then each decimal in the decimal representation of a/b is obtained by multiplying the remainder of the previous step by 10 and dividing it by b to get the new remainder. All remainders are between 0 and b − 1, so the process necessarily repeats itself.

Irrational numbers emerge quite naturally. The two most prominent examples are π half of the perimeter of the unit circle and 1 1 1 e 1 + + + +···. 1! 2! 3!

4A special case, known as one of Zeno’s paradoxes, can be explained to a first grader as follows: Stay 2 yards away from the wall. The goal is to reach the wall in infinite steps, in each step traversing half of the distance made in the previous step. Thus, in the first step you cover 1 yard, in the second 1/2, etc. You see that 1 + 1/2 + 1/22 +··· 2 1/(1 − 1/2). Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

2. “... There Are No Irrational Numbers at All”—Kronecker 11 e is also the principal and interest of $1 in continuous compounding after 1 year with 100% interest rate: n 1 e lim 1 + . n→∞ n

The equivalence of the two definitions of e is usually proved5 in calculus. ♦ To prove that π and e are irrational, we follow Hermite’s argument, which dates back to 1873. Consider, for fixed n ∈ N, the function f : R → R defined by xn( − x)n 2n f(x) 1 1 c xk. n n k ! ! k n n Expanding (1 − x) , we see that ck ∈ Z, k n, n + 1,...,2n.In c (− )k n (− )k n fact, k 1 n−k 1 k (by the binomial formula), but we will not need this. For 0 2n (m) f (0) cmm!  , if n ≤ m ≤ 2n. n! In any case f (m)(0) ∈ Z (since n! divides m!forn ≤ m). Thus, f and all its derivatives have integral values at x 0. Since f(x) f(1−x), the same is true for f at x 1.

Theorem 1. ek is irrational for all k ∈ N.

Proof. ¬ Assume that ek a/b for some a, b ∈ Z. Deﬁne F : R → R by F(x) k2nf(x) − k2n−1f (x) + k2n−2f (x) +···−kf (2n−1)(x) + f (2n)(x).

5For a direct proof based on the binomial formula (with integral coefﬁcients), see G.H. Hardy, A Course of Pure Mathematics, Cambridge University Press, 1960. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

12 2. “... There Are No Irrational Numbers at All”—Kronecker

The sum F (x) + kF(x) is telescopic; in fact, we have F + kF(x) k2n+1f(x).

We now use a standard trick in differential equations (due to Bernoulli) and multiply both sides by the integrating factor ekx to obtain [ekxF(x)] ekx[F (x) + kF(x)] k2n+1ekxf(x). Integrating both sides between 0 and 1:

1 k2n+1 ekxf(x) dx ekxF(x) 1 ekF( ) − F( ) [ ]0 1 0 0 (1/b)(aF(1) − bF(0)). In other words,

1 bk2n+1 ekxf(x) dx aF(1) − bF(0) 0 is an integer. On the other hand, since f(x) < 1/n!, 0

1 bk2n+1ek (k2)n 0

Remark. Another proof of irrationality of e can be given using the geometric series formula as follows: Let [·]:R → Z be the greatest integer function deﬁned by [r] the greatest integer ≤ r. Let k ∈ N.We claim that k 1 [k!e] k! . j j 0 ! Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

2. “... There Are No Irrational Numbers at All”—Kronecker 13

Indeed, since the left-hand side is an integer, all we have to show is ∞ k 1 < . ! j 1 j k+1 ! For this, we estimate ∞ k 1 ! j j k+1 ! k k k ! + ! + ! +··· (k + 1)! (k + 2)! (k + 3)!

1 + 1 + 1 +··· k + 1 (k + 1)(k + 2) (k + 1)(k + 2)(k + 3) 1 1 1 1 < + + +··· − 1 1. 2 22 23 1 − (1/2)

k k /j

Corollary. eq is irrational for all 0 q ∈ Q .

Proof. If eq is rational, then so is any power (eq)m eqm. Now choose m to be the denominator of q (written as a fraction) to get a contradiction to Theorem 1.

Remark. Q is dense in R in the sense that, given any real number r, we can ﬁnd a rational number arbitrarily close to r. This is clear if we write r in a decimal representation and chop off√ the tail arbitrarily far away from the decimal point. For example, 2 1.414213562 ... means that the rationals 1, 1.4 14/10, 1.41 141/100, 1.414 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

14 2. “... There Are No Irrational Numbers at All”—Kronecker √ 1414/1000 ...get closer and closer to 2. Indeed, we have 1.42 1.96, 1.412 1.9981, 1.4142 1.999396.

Theorem 2. π is irrational.6

Proof. It is enough to prove that π2 is irrational. ¬ Assume that π2 a/b for some a, b ∈ Z. Using the idea of the previous proof, we deﬁne F : R → R by F(x) bn[π2nf(x) − π2n−2f (x) + π2n−4f (4)(x) −···+(−1)nf (2n)(x)], where f is given above. Since π2n an/bn, F(0) and F(1) are again integers. This time, F (x) + π2F(x) is telescopic, so that we have [F (x) sin(πx) − πF(x) cos(πx)] [F (x) + π2F(x)] sin(πx) bnπ2n+2f(x) sin(πx) anπ2f(x) sin(πx). Integrating, we obtain

1 1 anπf(x) (πx) dx F (x) (πx) − πF(x) (πx) 1 sin [ sin cos ]0 0 π F(1) + F(0) ∈ Z. On the other hand,

1 anπ 0 < anπf(x) sin(πx) dx < < 1 0 n! if n is large enough. ¬

♣ Although π is irrational, it is not only tempting but very important to ﬁnd good approximations of π by rational numbers7 such as the Babylonian 25/8 (a clay tablet found near Susa) or the Egyptian

6Irrationality of π was ﬁrst proved by Lambert in 1766. 7For an interesting account on π, see P. Beckmann, A History of π, The Golem Press, 1971. For a recent comprehensive treatment of π, see L. Berggren, J. Borwein, and P. Borwein, π: A Source Book, Springer, 1997. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

2. “... There Are No Irrational Numbers at All”—Kronecker 15

28/34 (Rhind papyrus, before 1650 b.c.). The Babylonians obtained the ﬁrst value by stating that the ratio of the perimeter of a regular hexagon to the circumference of the circumscribed circle was “equal” to 57/60 + 36/602. (Recall that the Babylonians used 60 as a base.) The Egyptians obtained the second value by inscribing a circle into a square of side length 9 units, putting a 3 × 3 grid on the square by trisecting each edge, and approximating the circle by the octagon obtained by cutting off the corner triangles (Figure 2.1). The area of the octagon, 63 units, was rounded up to 82, and the approximate value of π turned out to be 82/(9/2)2 28/34. A well-known recorded approximation of π was made by Archimedes (c. 287–212 b.c.), who applied Eudoxus’s method of exhaustion, the approximation of the circumference of the unit circle by the perimeters of inscribed and circumscribed regular polygons. By working out the perimeters of regular 96-sided polygons inscribed in and circumscribed about a given circle, he established the estimate 10 1 3 <π<3 . 71 7 The difference between the upper and lower bounds is 1/497 ≈ 0.002, a remarkable accomplishment! To arrive at this estimate, Archimedes ﬁrst worked out the perimeters of the inscribed and

Figure 2.1 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

16 2. “... There Are No Irrational Numbers at All”—Kronecker

circumscribed hexagons and obtained 6 3 <π< √ . 3 √ He then estimated 3 from below by 265/153 (4-decimal precision!) and obtained 918/265 as an upper bound. Finally, to arrive at a polygon with 96 6 × 24 sides, he progressively doubled the sides of the polygons and made√ more and more reﬁned estimates of certain radicals (such as 349450 591 1/8). The algorithm Archimedes used is treated in Problem 12. Other rational approximations of π are 22√/7 (3-decimal precision) and 355/113 (6-decimal precision). 10 was used by Brahmagupta (c. a.d. 600). Numerous ingenious approximations of π were given by Ramanujan, such as the 9-decimal precision √ + 63 17 15√ 5 . 25 7 + 15 5 and the inﬁnite series expansion √ ∞ ( j) ( + j) 1 2 2 4 ! 1103 26390 , π (j )4 4j 9801 j 0 ! 396 where each term of the series produces an additional 8 correct digits in the decimal representation of π. ♦ Another way to approximate π is to use the integral formula

y dt tan−1 y 2 0 1 + t from calculus, expand the integrand into a power series and integrate. In fact, replacing x by −t2 in the geometric series formula, we obtain 1 1 − t2 + t4 − t6 + t8 − t10 +···, 1 + t2 so that

y dt y3 y5 y7 y9 y11 tan−1 y y − + − + − +···, 2 0 1 + t 3 5 7 9 11 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

2. “... There Are No Irrational Numbers at All”—Kronecker 17 and, setting y 1, we arrive at the Gregory–Leibniz series π 1 1 1 1 1 1 − + − + − +···. 4 3 5 7 9 11 (The Scottish mathematician David Gregory discovered the inﬁ- nite series expansion of tan−1 in 1671, three years before Leibniz derived the alternating series for π/4. Published by Leibniz in 1682, the expansion is sometimes named after him. For Størmer’s ap- −1 proach to the Gregory numbers ta/b tan (b/a), a, b ∈ N, see Problem 14.) Since the series on the right-hand side converges very slowly (several hundred terms are needed for even 2-decimal precision!), numerous modiﬁcations of this procedure have been made, using, for example,

π/4 4tan−1(1/5) − tan−1(1/239)

(see Problem 14) and

π/4 6tan−1(1/8) + 2tan−1(1/57) + tan−1(1/239).

To close this section, we give a brief account on a geometric method devised by Gauss that produces rational approximations of π with arbitrary decimal precision. Consider the two-dimensional square lattice Z2 ⊂ R2 of points with integral coordinates. For r>0, let N(r) denote the total number of lattice points contained in the ¯ (closed) disk Dr with center at the origin and radius r:

¯ 2 N(r) |Dr ∩ Z | ¯ (see Figure 2.2). Each lattice point (a, b) ∈ Dr, contributing to N(r), can be considered as the lower left corner of a unit square with sides parallel to the coordinate axes. Thus, N(r) can be thought of as the total area of the squares whose lower left corner is contained ¯ 2 ¯ in Dr. To see how far N(r) is from the area πr of Dr, let B(r) denote the total area of those squares that intersect the boundary circle of ¯ 2 Dr. We have |N(r) − πr √|≤B(r). Moreover, since the diagonal of a unit square has length 2, √ √ √ B(r) < π((r + 2)2 − (r − 2)2) 4 2πr. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

18 2. “... There Are No Irrational Numbers at All”—Kronecker

Figure 2.2

Dividing by r2, we obtain the estimate √ N(r) π − π < 4 2 . r2 r

Choosing r to be an integer, we see√ that N(r)/r2 gives a rational approximation of π with precision 4 2π/r. For example, r 300 gives N(300)/3002 3.14107 (3-decimal precision). ♥ To see an interesting connection with number theory, notice that N(r) counts the number of ways that nonnegative integers n ≤ r2 can be written as sums of two squares of integers. As shown in number theory,8 the number of representations of a positive integer n as a sum of two squares is four times the excess in the number of divisors of n of the form 4m + 1 over those of the form 4m + 3 (cf. also the discussion at the end of Section 1). In terms of the greatest integer function, the total number of divisors of the form 4m + 1 of all positive integers n ≤ r2 is

r2 r2 [r2] + + +···, 5 9

8See I. Niven, H. Zuckerman, and H. Montgomery, An Introduction to the Theory of Numbers, Wiley, 1991. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

2. “... There Are No Irrational Numbers at All”—Kronecker 19 and those of the form 4m + 3is

r2 r2 r2 + + +···. 3 7 11

Taking into account the trivial case (0 02 + 02), we obtain

N(r) − 1 r2 r2 r2 r2 r2 [r2] − + − + − +···. 4 3 5 7 9 11

The right-hand side contains only finitely many nonzero terms. Assuming from now on that r is an odd integer, we find that the number of nonzero terms is (r2 + 1)/2. To estimate this alternating series, we remove the greatest integer function from the first r terms and observe that we make an error less than r. The remaining terms are dominated by the leading term [r2/(r + 1)] since the series is alternating. We thus have N(r) − 1 r2 r2 r2 r2 r2 r2 r2 − + − + − + ...± ± cr, 4 3 5 7 9 11 r where 0 ≤ c ≤ 2. Dividing by r2 and letting r →∞through odd integers, we recover the Gregory–Leibniz series π 1 1 1 1 1 1 − + − + − +···. 4 3 5 7 9 11 ♦ For the approximation of π above we used unit squares with 2 ¯ vertices in Z to more or less cover the closed disk Dr for r large. Instead of squares we can use parallelograms Fa,b F + (a, b), a, b ∈ Z (Fa,b is F translated by the vector (a, b)), where the fun- 2 damental parallelogram F F0,0 (and hence Fa,b) has vertices in Z but no other points in Z2 (not even on the boundary). (See Figure 2.3.) The parallelograms Fa,b tesselate the whole plane in the sense that they cover R2 with no overlapping interiors. Tomake F unique we can assume that a specified vertex of F is the origin. The two vertices of F adjacent to the origin are then given by vectors v and w (with integral components), and the vertex of F opposite to the origin is v + w. The argument of Gauss goes through with obvious modifications. We obtain

2 2 2 |NF (r) − πr |

20 2. “... There Are No Irrational Numbers at All”—Kronecker

Figure 2.3

where δ is the diameter, the length of the longer diagonal, of F; NF (r) is the total area of the parallelograms whose speciﬁed vertex ¯ lies in Dr; and BF (r) is the total area of the parallelograms that ¯ intersect the boundary circle of Dr. Since NF (r) area (F)N(r), letting r →∞, we obtain

NF (r) N(r) lim area (F) lim . r→∞ r2 r→∞ r2 Since both limits are equal to π, we conclude that the area of a fundamental parallelogram in a square lattice is always unity. Since F is fundamental, the integral linear combinations of v and w exhaust Z2; i.e., we have Z2 {kv + lw | k, l ∈ Z}. The passage from the unit square to the fundamental parallelogram F is best described by the transfer matrix A between the bases {(1, 0), (0, 1)} and {v, w}. The column vectors of A are v and w.In particular, A has integral entries. Viewing v and w as vectors in R3 with zero third coordinates, we have area (F) |v × w| |det (A)|. This is unity, since F is fundamental. We thus have det (A) ± 1. Switching v and w changes the sign of the determinant. A 2 × 2 matrix with integral entries and determinant 1 (and the associ- Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 21 ated linear transformation) is called unimodular. The inverse of a unimodular matrix is unimodular, and the product of unimodular matrices is also unimodular. ♥ Thus, under multiplication, the unimodular matrices form a group. This is called the modular group ab ( , ) A | (A) ad − bc , a,b,c,d∈ . SL 2 Z cd det 1 Z

The unimodular transformations preserve the square lattice Z2 (in fact, this property can be used to define them). Finally, the unimodular transformations are area-preserving. The proof of this is an application of the argument of Gauss above to planar sets more general than disks. The only technical difficulty is to define what area means in this general context.

Problems √ 1. Let a be a positive integer such that n a is rational. Show that a must be the nth power of an integer. 2. ♦ Show that cos 1 is irrational. 3. Show that the cubic polynomial P(x) 8x3 − 6x − 1 has no rational roots. Use the trigonometric identity cos(θ) 4 cos3(θ/3) − 3 cos(θ/3) to conclude that cos 20◦ is a root of P, and thereby irrational. ♦ a xn dx a> n − 4. Work out the integral 1 , 1, 1, using the geometric series formula in the following way: Let m ∈ N and consider the Riemann√ sum {αk}m α m a (approximating the integral) on the partition k 0, where . 5. ♦ Show that, for n ∈ N, we have n(n + 1) 1 + 2 +···+n . 2 6. Solve Problem 5 in the following geometric way: Consider the function f : [0,n] → R, deﬁned by f(x) 1 + [x], 0 ≤ x ≤ n, and realize that the area of the region S under the graph of f gives 1+2+···+n.(S looks like a “staircase.”) Cut S into triangles and work out the area of the pieces. Generalize this to obtain the sum of an arithmetic progression. 7. ♦ Show that, for n ∈ N, we have n(n + 1)(2n + 1) 12 + 22 +···+n2 . 6 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

22 2. “... There Are No Irrational Numbers at All”—Kronecker

8. Solve Problem 7 along the lines of Problem 6 by building a 3-dimensional pyramid staircase whose volume represents 12 +22 +···+n2. Cut the staircase into square pyramids and triangular prisms. 9. Show that n(n + 1) 2 13 + 23 +···+n3 2 in the following geometric way (this was known to the Arabs about 1000 years ago): For k 1,...,n, consider the square Sk ⊂ R2 with vertices (0, 0), (k(k + 1)/2, 0), (k(k + 1)/2,k(k+ 1)/2), and (0,k(k+ 1)/2).Forn ≥ 2, view Sn as the union of S1 and the L-shaped regions Sk − Sk−1, k 2,...,n. By splitting Sk − Sk−1 into two rectangles, show that the area of Sk − Sk−1 is k3.

10. Given n ∈ N, define a finite sequence of positive integers: a0 a, a1 [a0/2], a2 [a1/2],.... Show that for b ∈ N we have ab 2nb. an odd (This algorithm of multiplication, consisting of systematic halving and doubling, was used by the ancient Greeks, and was essentially known to the ancient Egyptians.) 11. A triangular array of dots consists of 1, 2,...,n dots, n ∈ N, stacked up to form a triangle. The total number of dots in a triangular array is the nth triangular number Tri (n) 1 + 2 + 3 +···+n n(n + 1)/2 (cf. Problem 5). (a) Define the nth square number Squ (n) n2, n ∈ N. By fitting two triangular arrays into a square, show that Squ (n) Tri (n) + Tri (n − 1). (This is due to Nicomachus, c. a.d. 100.) Define the nth pentagonal number Pen (n) Squ (n) + Tri (n − 1) and arrange the corresponding dots in a pentagonal array. Generalize this to define all polygonal numbers. (b) By fitting 8 triangular arrays into a square, show that a ∈ N is a triangular number iff 8a + 1 is a perfect square. (This is attributed to Plutarch, c. a.d. 100.) (c) Prove that ∞ 1 . (n) 2 n 1 Tri (d) Define the nth k-gonal number inductively9 as the sum of the nth (k−1)- gonal number and Tri(n − 1). Show that the nth k-gonal number is equal to ((k − 2)n2 − (k − 4)n)/2 by observing that this number is the sum of the first n terms of an arithmetic progression with first term 1 and common difference k − 2. Conclude that a hexagonal number is triangular, and a pentagonal

9See L.E. Dickson, History of the Theory of Numbers, II, Chelsea Publishing, 1971. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 23

Figure 2.4

number is 1/3 of a triangular number. What is the geometry behind these conclusions?

12. Let sn and Sn denote the side lengths of the inscribed and circumscribed regular n-sided polygons to a circle. (a) Derive the formulas S s 2n n s2 s S and 2 2n n 2n Sn − S2n Sn by looking for similar triangles in Figure 2.4. (b) Let pn nsn and Pn nSn be the corresponding perimeters. Use (a) to conclude that 2Pnpn P2n and p2n pnP2n. Pn + pn (c) Assume that the circle has unit radius and show that √ √ √ p6 6,P6 4 3,p12 12 2 − 3,P12 24 2 − 3 .

How well do p12 and P12 approximate 2π? (d) Use the inequality ab √ 2 ≤ ab, a, b > , a + b 0 to show10 that P p2

10 This interpolation technique is due to Heinrich Dorrie,¨ 1940. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

24 2. “... There Are No Irrational Numbers at All”—Kronecker

Deﬁne Pnpn 3 2 3 An Pnpn and Bn . 2Pn + pn

Translate the inequalities above to prove monotonicity of the sequences {An} and {Bn}, and conclude that

Bn < 2π

How well do A12 and B12 approximate 2π? Does this method surpass Archimedes’ approximation of π using 96-sided polygons? √ 13. Evaluate the expansion of tan−1 at 1/ 3, and obtain π 1 1 1 1 √ 1 − + − +··· . 6 3 3 · 3 32 · 5 33 · 7 (This expansion was used by Sharp (1651–1742) to calculate π with 72 decimal precision.)

14. Deﬁne the rth Gregory number tr , r ∈ R, as the angle (in radians) in an uphill − road that has slope 1/r. Equivalently, let tr tan 1(1/r). (As indicated in the text, Gregory found the inﬁnite series expansion 1 1 1 tr − + −···, r 3r3 5r5 an equivalent form of the Taylor series for tan−1.) ♦ Use Størmer’s observa- tion11 that the argument of the complex number a + bi is ta/b (and additivity of the arguments in complex multiplication) to derive Euler’s formulas

t1 t2 + t3,

t1 2t3 + t7,

t1 5t7 + 2t18 − 2t57. Generalize the ﬁrst equation to prove Lewis Caroll’s identity 2 tn tn+s + tn+t iff st n + 1, and Machin’s formula

t1 4t5 − t239. The latter translates into π/4 4tan−1(1/5) − tan−1(1/239).

11 For more details, see J. Conway and R. Guy, The Book of Numbers, Springer, 1996. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Web Sites 25

(This was used by Machin, who calculated π with 100-decimal precision in 1706.) Derive the last formula using trigonometry. (Let tan θ 1/5 and use the double angle formula for the tangent to work out tan(2θ) 5/12 and tan(4θ) 120/119. Notice that this differs from tan(π/4) 1by1/119. Work out the error tan(4θ − π/4) 1/239 using a trigonometric formula for the tangent.) 15. ♦ Given a basis {v, w} in R2, the set L {kv + lw | k, l ∈ Z} is called a lattice in R2 with generators v and w. Generalize the concept of fundamental parallelogram (from Z2)toL, and show that every possible fundamental parallelogram of a lattice L has the same area. Use this to prove that the transfer matrix, with respect to the basis {v, w}, between two fundamental parallelograms is unimodular.

Web Sites

1. www.maa.org 2. www.math.psu.edu/dna/graphics.html 3. www.dgp.utoronto.ca/people/mooncake/thesis 4. www.mathsoft.com/asolve/constant/pi/pi.html Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

. . . . . Rationality, Elliptic . . . . . Curves, and Fermat’s . . . 3 . SECTION . Last Theorem

♣ A point in R2 with rational coordinates is called rational. The rational points Q 2 form a dense set in R2 (that is, any open disk in R2 contains a point in Q 2). Theorem 1 of Section 2 can be reformulated by saying that the only rational point on the graph

{(x, ex) | x ∈ R}⊂R2

of the exponential function is (0, 1). (It is quite amazing that this smooth curve misses all points of the dense subset Q 2 −{(0, 1)} of R2!) An easier example (to be generalized later to Fermat’s famous Last Theorem) is the unit circle

S1 {(x, y) ∈ R2 | x2 + y2 1},

and we ask the same question of rationality: What are the rational points on the unit circle? Apart from the trivial (±1, 0), (0, ±1), and the less trivial examples (3/5, 4/5), (5/13, 12/13), and (8/17, 15/17), we can actually give a complete description of these points. Indeed, assume that x and y are rational and satisfy x2 + y2 1. Write x and y as fractions of relatively prime integers x a/c and y b/d. Substituting and multiplying out both sides 26 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 27 of the equation by c2d2, we obtain a2d2 + b2c2 c2d2.

This immediately tells us that c2 divides a2d2, and d2 divides b2c2. Since a, c and b, d have no common divisors, c2 and d2 divide each other, and so they must be equal. The equation above reduces to a2 + b2 c2.

We see that the question of rationality is equivalent to ﬁnding positive integer solutions of the Pythagorean equation above. A solution (a,b,c)with a, b, c ∈ N is called a Pythagorean triple. The obvious reason for this name is that each Pythagorean triple gives a right triangle with integral side lengths a, b, and c (with c corresponding to the hypotenuse). By the way, in case you have not seen a proof of the Pythagorean Theorem,1 Figure 3.1 depicts one due to Bhaskara (1114–c. 1185). The rotated 4-sided polygon on the right is a square, which you can show from symmetry or by calculating angles.

Remark. Another cutting-and-pasting gem of Greek geometry is due to Hip- pocrates of Chios (c. 430 b.c.), and it states that the crescent lune in Figure 3.2 has the same area as the inscribed triangle. To show this, verify that A 2a in Figure 3.3. The signiﬁcance of this should not be underrated. In contrast to the futile attempts to

c c

a b a b Figure 3.1

1Have you noticed that the Pythagorean Theorem, the Euclidean distance formula, the trigonometric identity sin2 θ + cos2 θ 1, and the equation of the circle are all equivalent to each other? Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

28 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

Figure 3.2

square the circle, this is the ﬁrst recorded successful measurement of the exact area of a plane ﬁgure bounded by curves! All Pythagorean triples have been known since the time of Eu- clid (see Book X of Euclid’s Elements2) and can be found in the third century work Arithmetica by Diophantus. An ancient Babylonian tablet (c. 1600 b.c.) contains a list of Pythagorean triples including (4961, 6480, 8161)! It indicates that the Babylonians in the time of

a a

Figure 3.3

2T.L. Heath, The Thirteen Books of Euclid’s Elements, Dover, New York, 1956. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 29

Hammurabi were well aware of the significance of these numbers, including the fact that they are integral side lengths of a right triangle. The time scale is quite stunning; this is about 1000 years before Pythagoras! The first infinite sequence of all Pythagorean triples (a,b,c)with b,c consecutive was found by the Pythagoreans themselves: (a,b,c) (n, (n2 − 1)/2,(n2 + 1)/2), where n>1 is odd. Plato found Pythagorean triples (a,b,c)with b + 2 c, namely, (4n, 4n2 − 1, 4n2 + 1), n ∈ N. Finally, note also that Euclid, although he obtained all Pythagorean triples, provided no proof that his method gave them all. We give now a brief account of the solution to the problem of Pythagorean triples. We may assume that a, b, and c are relatively prime (i.e., a, b, and c have no common divisor). This is because (a,b,c) is a Pythagorean triple iff (ka, kb, kc) is a Pythagorean triple for any k ∈ N. (Notice also that (ka, kb, kc) gives the same rational point on S1 for all k ∈ N.) It follows that a and b are also relatively prime. (¬ If a prime p divides both a and b, then it also divides c2 a2 +b2. Being a prime, p also divides c, so that p is a common divisor of a, b, and c, contrary to our choice. ¬) Similarly, b, c and a, c are also relatively prime. We now claim that a and b must have different parity. Since they are relatively prime, they cannot be both even. ¬ Assume that they are both odd, say, a 2k + 1 and b 2l + 1. Substituting, we have c2 a2 + b2 4(k2 + k + l2 + l) + 2 2(2(k2 + k + l2 + l) + 1), but this is impossible since the square of a number cannot be the double of an odd number! ¬ (Compare this√ with the proof of Proposition 1 which shows the irrationality of 2.) Without loss of generality, we may assume that a is even and b is odd. Hence, c must be odd. Since the difference and sum of odd numbers is even, c − b and c + b are both even. We can thus write c − b 2u and c + b 2v, where u, v ∈ N. Equivalently, c v + u and b v − u. These equations show that u and v are relatively prime (since b and c are). An argument similar to the one above shows that u and v have different parity. Recall now that a is even, so that a/2isan Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

30 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

integer. In fact, in terms of u and v, we have a 2 c − b c + b uv. 2 2 2 Let p be a prime divisor of a/2. Then p2 divides uv; and since u and v are relatively prime, p2 must divide either u or v. It follows that u and v are pure square numbers: u s2 and v t2. In addition, s and t are also relatively√ prime and of different parity. With these, we have a 2 uv 2st. Putting everything together, we arrive at a 2st, b t2 − s2,c t2 + s2. We claim that this is the general form of a Pythagorean triple (with relatively prime components) parametrized by two integers s and t, where t>s, and s, t are relatively prime and of different parity. To show this we just have to work backward: a2 + b2 4s2t2 + (t2 − s2)2 (t2 + s2)2 c2, and we are done.

The following table shows a few values:

tsa 2st b t2 − s2 c t2 + s2 21 4 3 5 3 2 12 5 13 41 8 15 17 4 3 24 7 25 5 2 20 21 29 5 4 40 9 41 6 1 12 35 37

There is a lot of geometry behind the Pythagorean triples. For example, one can easily show that the radius of the inscribed circle of a right triangle with integral sides is always an integer (see Problem 2). On the number theoretical side, the components of a Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 31

Pythagorean triple (a,b,c)contain many divisors; for example, the Babylonian 60 always divides the product abc! (See Problem 3.)

Remark. A much harder ancient problem is to ﬁnd a simple test to determine whether a given positive rational number r is a congruent number, the area of a right triangle with rational side lengths a, b, and c.In general, any nonzero rational number r can be made a square free integer by multiplying r by the square of another rational number s. As the name suggests, square free means that the integer (in this case s2r) has no nontrivial square integral divisors. Now, if r is the area of a triangle with side lengths a, b, c, then s2r is the area of a similar triangle with side lengths sa, sb, sc. Thus, we may (and will) assume that a congruent number is a square free natural number. As an example, we immediately know that 6 is a congruent number, since it is the area of a right triangle with side lengths 3, 4, 5. Euler showed that 7 is a congruent number, and Fermat that 1 is not. (The latter is a somewhat stronger form of Fermat’s last theorem in the exponent 4, to be treated later in this section; see Problem 20.) As a further example, 5 is a congruent number, as shown in Figure 3.4. Eventually, it became clear that 1, 2, 3, 4 are not congruent numbers, while 5, 6, 7 are. A major breakthrough in the “congruent number problem” was provided by Tunnell in 1983. His deep result, in its simplest form, states that if an odd square free natural number n is congruent, then the number of triples of integers (x,y,z)satisfying 2x2 + y2 + 8z2 n is equal to twice the number of triples of integers (x,y,z)satisfying 2x2 + y2 + 32z2 n.Forn an even square free natural number, the same is true with 2x2 replaced by 4x2, and n replaced by n/2. The converse statements are also true, provided that the so-called

b a=3/2, b=20/3, c=41/6 Figure 3.4 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

32 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

a = 6803298487826435051217540 411340519227716149383203

411340519227716149383203 b = 21666555693714761309610

224403517704336969924557513090674863160948472041 c = Figure 3.5 8912332268928859588025535178967163570016480830

Birch–Swinnerton–Dyer conjecture is true. To appreciate the subtlety of the problem, take a look at Figure 3.5, showing that 157 is a congruent number. (This is due to Zagier.)3

Returning to Pythagorean triples, we see that the rational points on the unit circle are of the form st t2 − s2 2 , t2 + s2 t2 + s2 where s, t are relatively prime integers. There is a beautiful geometric way to obtain these points. Con- sider a line l through the rational point (0, 1) ∈ S1 given by the equation y mx + 1. For m 0, l intersects S1 in a point other than (0, 1). This intersection point is obtained by coupling this equation with that of the

3Cf. N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer, 1993. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 33 unit circle. Substituting, we obtain x2 + (mx + 1)2 1. The left-hand side factors as x((1 + m2)x + 2m) 0, so that we obtain the coordinates of the intersection: 2m 1 − m2 x − and y mx + 1 . 1 + m2 1 + m2 This shows that (x, y) is a rational point on S1 iff m is rational. (Indeed, m ∈ Q clearly implies x, y ∈ Q . The converse follows from the equation m (y − 1)/x.) Varying m in Q , we obtain all rational points on S1 (except (0, −1), which corresponds to a vertical line intersection point). Setting m −s/t, t 0, s, t ∈ Z, we recover the solution set above. This “method of rational slopes” works for all quadratic curves described by f(x, y) 0, where f is a quadratic polynomial with rational coefficients in the variables x, y, provided that there is at least one rational point on the curve. There may not be any;√ for example, there are no rational points on the circle (with radius 3) given by the equation x2 + y2 3. This follows because we showed in Section 1 that 3 does not divide any sums of squares a2 +b2, a, b ∈ Z, with a and b relatively prime. We are now tempted to generalize this approach to find rational points on all algebraic curves. An algebraic curve is, by definition, the locus of points on R2 whose coordinates satisfy a polynomial equation f(x, y) 0. The locus as a point-set is denoted by

2 Cf {(x, y) ∈ R | f(x, y) 0}.

We say that Cf is rational if f has rational coefﬁcients. Among all the polynomials whose zero sets give the algebraic curve, there is one with minimum degree. This degree is called the degree of Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

34 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

the algebraic curve. We will consider cubic (degree-three) algebraic curves below. ♦ As in calculus, it is convenient to consider the graph

{(x,y,f(x,y)) ∈ R3 | (x, y) ∈ R2}

of the polynomial f . This is a smooth surface in R3. The algebraic curve Cf is obtained by intersecting this graph with the horizontal plane spanned by the ﬁrst two coordinate axes. (This is usually called the level curve of f corresponding to height 0.) We say that (x0,y0) is a critical point of f if the plane tangent to the graph at (x0,y0) is horizontal. In terms of partial derivatives, this means that ∂f ∂f (x ,y ) (x ,y ) . ∂x 0 0 ∂y 0 0 0

If (x0,y0) ∈ Cf is not a critical point of f , then near (x0,y0), Cf is given by a smooth curve across (x0,y0).IfCf is cubic and (x0,y0) ∈ Cf is a critical point of f , then near (x0,y0) the curve Cf displays essentially two kinds of singular behavior. At the assumed singular point, Cf either forms a cusp or it self-intersects in a double point (cf. Figure 3.6). The two typical nonsingular cubic curves are depicted in Figure 3.7. At present, we are most interested in the second type of singular point. This is the case when the singular point (x0,y0) of Cf is a saddle point for f . For example, compare the simple saddle point in Color Plate 1a depicting the graph of a quadratic polynomial, with Color Plate 1b, where the saddle is given by a cubic polynomial.

Figure 3.6 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 35

Figure 3.7

Again by calculus, this configuration is guaranteed by the “saddle condition” ∂2f ∂2f ∂2f 2 − < 0 ∂x2 ∂y2 ∂x∂y near (x0,y0). Assuming this, if (x0,y0) is on Cf , then intersecting the surface with the horizontal coordinate plane, we obtain that, near (x0,y0), Cf consists of two smooth curves crossing over at (x0,y0). In this case, we say that the algebraic curve Cf has a double point at (x0,y0). The significance of this concept follows from the fact that any double point on a cubic rational curve is rational! We will not prove this in general. ♣ Instead, we restrict ourselves to the most prominent class of cubic rational curves given by the so-called Weierstrass form y2 P(x), where P is a cubic polynomial with rational coefficients. (♠ As far as rationality is concerned, restricting ourselves to Weierstrass forms results in no loss of generality. In fact, any cubic curve is “birationally equivalent” to a cubic in Weierstrass form. This means that for any cubic curve a new coordinate system can be introduced; the new coordinates depend rationally on the old ones; and in terms of the new coordinates, the cubic is given by a Weierstrass form. Because of the rational dependence of the old and new coordinates, the rational points on the original curve correspond to rational points on the new curve. (For a special case, see Problem 19.) Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

36 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

Remark. ♣ The origins of this equation go back to ancient times. This is easily understood if we consider the special case P(x) x3 + c, c ∈ N, and realize that integer solutions of this equation are nothing but the possible ways to write c as the difference of a square and a cube:

y2 − x3 c.

This special case is called the Bachet equation. The Bachet curve corresponding to the Bachet equation is nonsingular unless c 0, in which case it reduces to a cusp. In 1917, Thue proved that for any c, there are only ﬁnitely many integral solutions to this equation. There may be none; for example, Bachet’s equation has no solutions for c 7. Although the proof is elementary, we will not show this, except to mention that the proof depends on writing the equation as

y2 + 1 x3 + 8 (x + 2)(x2 − 2x + 4)

and studying the remainders of x and y under division by 4. How- ever, Bachet’s equation most often has inﬁnitely many rational solutions. A little more about this later. We now return to cubics in Weierstrass form. A good example on which to elaborate is given by the equation

y2 x3 − 3x + 2.

The cubic curve is depicted in Figure 3.8. Here, (1, 0) is a double point. (This will follow shortly.) Color Plate 1b shows the intersection of the graph of the polynomial f(x, y) x3 − 3x + 2 − y2 in R3 with equidistant (horizontal) planes. Projecting the intersection curves to the coordinate plane spanned by the ﬁrst two axes, we obtain the level curves of f (Figure 3.9). This indeed shows the three basic types discussed earlier! ♦ Let us try our machinery for the Weierstrass form. Since f(x, y) P(x) − y2, we have ∂f ∂f P(x) − y. ∂x and ∂y 2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 37

Figure 3.8

Let (x0,y0) be a critical point of f . Then y0 0 and P (x0) 0. If, (x ,y ) ∈ C y2 P(x ) furthermore, 0 0 f , then 0 0 0. Thus, a critical point of f on Cf is of the form (r, 0), where P(r) P (r) 0. Differentiating once more, we see that (r, 0) is a saddle point if P(r) > 0. We now look at these conditions on P from an algebraic point of view. P(r) 0 means that r is a root of P. By the factor theorem, the root factor (x − r) divides P. The second condition P(r) 0 means that r is a root of P of multiplicity at least 2; that is, (x − r)2

Figure 3.9 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

38 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

divides P. Finally, P(r) > 0 means that r is a root of multiplicity exactly 2, or equivalently, (x − r)2 exactly divides P. Since P is of degree 3, we obtain the complete factorization: P(x) a(x − r)2(x − s), r s.

Since P has rational coefﬁcients, multiplying out, it follows by easy computation that r, and hence s, is rational! ♣ Summarizing, we ﬁnd that a double point on a cubic curve y2 P(x) occurs at (r, 0), where r ∈ Q is a root of multiplicity 2. In the example above, we see that y2 x3 −3x +2 (x−1)2(x+2), so that (1, 0) is a double point. (Moreover, it also follows that the cusp corresponds to a triple root of P; in fact, the cusp in Figure 3.6 is given by the singular Bachet equation y2 x3.) Returning to the general situation, we now apply the method of rational slopes to describe all rational points on the cubic curve y2 P(x) by considering lines through a rational double point (r, 0) whose existence we assume. The general form of a line through (r, 0) is given by y m(x − r).

Toﬁnd the intersection point, we put this and the deﬁning equation together and solve for x and y: m2(x − r)2 P(x).

Since r is a root of P with multiplicity two, P(x) a(x − r)2(x − s), r s,

where a, r, and hence s, are rational. Thus, x s + m2/a and y m(s − r + m2/a), and these are rational iff m is. As before, varying m on Q , we obtain all rational points on our cubic curve. In the elusive case where P has three distinct roots, the method of rational slopes does not work. A cubic rational curve in Weier- strass form y2 P(x), where the cubic polynomial P has no double or triple roots, is called elliptic.(♦ The name “elliptic” comes from the fact that when trying to determine the circumference of an ellipse one encounters elliptic integrals of the form R(x, y)dx, Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 39 where R is a rational function of the variables x and y, and y is the square root of a cubic or quartic polynomial in x.) By the discussion above, an elliptic curve is everywhere nonsingular.

Remark. The “congruent number problem” noted above is equivalent to a problem of rationality for a specific elliptic curve. As usual, we discuss only the beginning of this deep subject.4 Let n ∈ N be a congruent number. By definition, there exists a right triangle with rational side lengths a, b, c, and area n. We have a2 + b2 c2 and ab 2n. Adding or subtracting twice the second equation from the first, we get (a ± b)2 c2 ± 4n. Setting x (c/2)2,we see that x, x + n, and x − n are squares of rational numbers. We obtain that if a natural number n is congruent, then there exists a nonzero rational number x such that x, x + n, and x − n are squares of rational√ numbers.√ The converse√ of this statement√ is also a true,√ since x + n − x − n, b x + n + x − n, and c 2 x are the rational side lengths of a right triangle with area n, provided that x is a nonzero rational number with x, x +n, and x −n squares of rational numbers. If a, b, c are the rational side lengths of a right triangle with congruent area n, then multiplying the two equations ((a±b)/2)2 (c/2)2 ±n together, we get ((a2 −b2)/4)2 (c/2)4 − n2. We see that the equation u4 − n2 v2 has a rational solution u c/2 and v (a2 − b2)/4. Setting x u2 (c/2)2 and y uv (a2 − b2)c/8, we obtain that (x, y) is a rational point on the elliptic curve in the Weierstrass form

y2 x3 − n2x.

♣ The theory of elliptic curves displays one of the most beautiful interplays between number theory, algebra, and geometry. In what follows, all the examples of elliptic curves will be in Weierstrass form y2 P(x), where we assume that P has rational coefﬁcients and no multiple roots. (For an elliptic curve in Weierstrass form, P has either one or three real roots as in Figure 3.7.)

4See N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer, 1993. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

40 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

As a variation on the theme, we try the following “chord-method”: Given an elliptic curve, we consider the line l though two of its rational points, say, A and B. The slope of l is rational since A and B have rational coordinates. Since the elliptic curve is cubic, there must be a third intersection point with l. We denote this by A ∗ B. This point is rational. (This follows from the fact that if a cubic polynomial has rational coefficients and two rational roots, then the third root is also rational. Indeed, the sum r1 + r2 + r3 of the 2 3 three roots of a cubic rational polynomial c0 + c1x + c2x + c3x c3(x − r1)(x − r2)(x − r3), c3 0, is equal to −c2/c3 ∈ Q .) The operation (A, B) → A ∗ B on the set of rational points of an elliptic curve does not define a group structure, since there is no point O playing the role of the identity element. Instead, we fix a rational point O (assuming it exists) and define A + B as the third intersection of the line through O and A ∗ B. Before going further, take a look at Figure 3.10. Our aim is to show that the operation (A, B) → A + B defines a group structure5 on the set of rational points on the elliptic curve. Ignoring the finer point of what happens if the initial points co-

A ∗ B A B

A + B

Figure 3.10

5Although the deﬁnition of the sum of two rational points on an elliptic curve appears in the works of Cauchy (c. 1835), the fact that this operation deﬁnes a group structure was only recognized by Poincare´ around 1901. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 41

A∗O

A+O = A

Figure 3.11 incide, it is easy to give visual proofs that the group axioms are satisfied. We begin with the identity A + O A (Figure 3.11). For the existence of the negative of a point, the identity A + (−A) O requires that the line through A ∗ (−A) and O should have no further intersection with the elliptic curve. It follows that this line must be tangent to the curve at O. (As will be shown below, a nonvertical line tangent to the elliptic curve meets the curve at exactly one other point.) The construction of −A is depicted in Figure 3.12. Associativity A+(B+C) (A+B)+C is slightly more complex6 (Figure 3.13). Since commutativity (A + B B + A) is clear, we are done! Now, the finer point of coincidences. As we saw above, all is well and the chord construction applies if A B; but what happens if A B? At this point we have to rely on our geometric intuition. As usual, let the distinct A and B define A ∗ B as the third intersection point of the line l through A and B. Now pretend to be Newton and let B approach A. At the limit point A B, the chord l becomes tangent to the elliptic curve at A B! Thus, we define A ∗ A to be the intersection of the tangent at A with the elliptic curve. The big

6For details, see J. Silverman and J. Tate, Rational Points on Elliptic Curves, Springer, 1992. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

42 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

−A

Figure 3.12

question, of course, is whether or not this intersection is a single point (including the case when it is empty)!

Remark. For Bachet’s equation y2 x3 + c, c ∈ Z, this so-called tangent method gives an interesting duplication formula. Given a rational solution A ∈ Q 2, the line tangent to the algebraic curve at A

O A B C

B + C

A + B

(A + B) ∗ C = A ∗ (B + C) Figure 3.13 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 43

Figure 3.14

meets the curve at another point B that is easily seen to be rational (Figure 3.14). Letting A (x, y), y 0, computation shows that x4 − cx −x6 − cx3 + c2 B 8 , 20 8 . 4y2 8y3

By using ingenious algebraic manipulations, Bachet discovered this duplication formula in 1621, before calculus was developed (see Problem 9). Before going any further, let us play around with this formula in two special cases. Since 32 23 + 1, we see that A (2, 3) is an integer point on the Bachet curve given by y2 x3 + 1. The duplication formula gives A ∗ A A ∗ A ∗ A ∗ A (0, −1). (For more on this, see Problem 18.) For a less trivial example, start with 3 22 − 1, multiply both sides of this equation by 24 · 33, and obtain (22 · 32)2 (22 · 3)3 − 24 · 33. This shows that A (22 · 3, 22 · 32) (12, 36) is an integer point on the Bachet curve given by y2 x3 − 432. (If this is too ad hoc, take a look at Problem 19.) The duplication formula gives A ∗ A A! These two examples seem to crush our hope that by applying the duplication formula iteratively to an initial rational point we will obtain inﬁnitely many distinct rational points on our Bachet curve. Fortunately, these are Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

44 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

the only exceptional cases, and it can be proved7 that starting with A (x, y), with xy 0 and c 1, −432, we do get inﬁnitely many distinct rational points on the Bachet curve. ♦ Back to A ∗ A! Oveowing with conﬁdence, we now believe that we can handle this technically and actually compute A ∗ A. This needs a little calculus. Let A (x0,y0) be a rational point y2 P(x) y2 P(x ) on the elliptic curve in question; 0 0 . Assuming y0 0, the slope m y dy/dx of the tangent at x0 can be obtained by implicit differentiation. We have 2yy P(x),

which at x0 gives P(x ) m 0 ∈ Q . 2y0 (Note that the numerator is rational because P is.) Substituting this into the equation y − y0 m(x − x0), we obtain the equation of the tangent line P (x0) y y0 + (x − x0). 2y0 To look for intersections, we combine this with the deﬁning equation of the elliptic curve and obtain 2 P (x0) y0 + (x − x0) P(x). 2y0 Equivalently,

2 (2P(x0) + P (x0)(x − x0)) 4P(x0)P(x). Several terms look as if they jumped out of Taylor’s formula. In fact, expanding P into Taylor series, we have P(x ) P(x) P(x ) + P(x )(x − x ) + 0 (x − x )2 0 0 0 2 0 P(x ) + 0 (x − x )3. 6 0

7See N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer, 1993. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 45

(Notice that since P is cubic this is all we have.) After substituting, and allowing some mutual self-destruction of the opposite terms, we end up with

P(x )2 − P(x )P(x ) x x + 0 2 0 0 . 0 3 2P (x0)P(x0)

This gives the first rational coordinate of the intersection A ∗ A. (Do not worry about the denominator; it is twelve times the leading coefficient of P, and thereby nonzero.) Notice that, in the particular case P(x) x3 + c, our result reduces to Bachet’s duplication formula! Since the point (x0,y0) is on a line with rational slope, the second coordinate is also rational. We thus obtain a unique rational point A ∗ A. The remaining case y0 0 sets the tangent line vertical (perpendicular to the first axis) at x0 which, by assumption, P y2 P(x ) is a rational root of (since 0 0 0). Clearly, there is no further intersection of this line with the elliptic curve. The usual way to circumvent this difficulty is to attach “ideal points” to the Euclidean plane to each “direction” or, more precisely, to each pencil of parallel lines. Attaching these ideal points to the Euclidean plane, we obtain the classical model of the projective plane. More about this in Section 16. ♣ Here we just say that the vertical tangent line intersects the extended elliptic curve at the ideal point given by the pencil of vertical lines. Since the vertical ideal point has to be attached to our elliptic curve, we might as well choose it to be the identity O! Addition then takes a simple form depicted in Figure 3.15. The negative of a point is illustrated in Figure 3.16. (In fact, keep A fixed in Figure 3.15 and let B tend to −A. Then both A ∗ B and A + B tend to the ideal point.) Summarizing, we find that the rational points on a rational elliptic curve in the projective plane form a group under the addition rule (A, B) → A + B defined above. ♠ A deep theorem of Mordell (1923) asserts that this group is finitely generated. More plainly, this means (in the spirit of Ba- chet’s duplication formula) that on an elliptic curve with at least one rational point, there exist finitely many rational points such that all other rational points can be “generated” from these by the Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

46 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

A ∗ B B

A + B

Figure 3.15

“chord-and-tangent” method. This gives a very transparent description of the rational points on any elliptic curve. As an example, the rational points of the elliptic curve8 y2 + y x3 − x (y is shifted!) form an inﬁnite cyclic group (thereby isomorphic with Z). In Fig- ure 3.17, the zero O is set at the vertical ideal point and the rational points are labelled with the integers (isomorphically).

−A

Figure 3.16

8See R. Hartshorne, Algebraic Geometry, Springer, 1977. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 47

Figure 3.17. R. Hartshorne, Algebraic Geometry, 1977, 336. Reprinted by permission of Springer-Verlag New York, Inc.

Remark. We saw above that if n is a congruent number, then each right triangle with rational side lengths and area n can be made to correspond to a specific rational point on the elliptic curve En given in Weier- 2 3 2 9 strass form by y x − n x. It turns out that the points A on En, for n congruent, that correspond to “rational triangles” are exactly the doubles of rational points (A B + B 2B for B rational) on En. On the other hand, in the group of rational points of En, the only rational points of finite order are the four points of order two: the identity O (the vertical ideal point), (0, 0), and (±n, 0) (cf. Problem 11). In this statement we need only to assume that n is a square free natural number. If n is a congruent number, then the rational point that corresponds to the right triangle with side lengths a, b, c, and area n, has x-coordinate x (c/2)2 (see the computations above). Since this is different from the x-coordinates of the four points of order two above, we see that this rational point must be of infinite order. The converse of this statement is also true. We obtain that n is a congruent number iff the group of rational points of En is infinite! This is the first (and most elementary) step in proving Tunnell’s characterization of congruent numbers. All that we said here can be put in an elegant algebraic framework. If we denote

9See N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer, 1993. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

48 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

by En(Q ) the group of rational points on En, then, by Mordell’s theorem, En(Q ) is finitely generated. Since this is an abelian group, its elements of finite order form a (finite) subgroup, the so-called torsion subgroup: En(Q )tor. By what we said above, the order of this torsion subgroup is 4. By the fundamental theorem on finitely generated abelian groups, ∼ r En(Q ) En(Q )tor × Z ,

where r is called the rank of En(Q ). Again by the above, En(Q ) has nonzero rank iff n is a congruent number. It turns out that determining r is much more difﬁcult than locating the rational points of ﬁnite order on En.

♣ Another direction in which to generalize the Pythagorean problem is to ﬁnd rational points on the algebraic curve {(x, y) ∈ R2 | xn + yn 1} for n ≥ 3. The graph of x4 + y4 1 is depicted in Figure 3.18. The method of rational slopes breaks down, despite the fact that (1, 0) and (0, 1) are rational points. (Try to pursue this for n 3.) As in the Pythagorean case, we can reformulate this problem to ﬁnding all positive integer solutions of the equation an + bn cn,n≥ 3. This problem goes back to Fermat10 (1601–1665), who wrote the following marginal note in his copy of Diophantus’s Arithmetica: It is impossible to write a cube as a sum of two cubes, a fourth power as a sum of fourth powers, and, in general, any power beyond the second as a sum of two similar powers. For this, I have discovered a truly wonderful proof, but the margin is too small to contain it. Fermat thus claimed that there is no all-positive solution of this equation for any n ≥ 3. His “truly wonderful proof” went with him to the grave, and despite intense efforts of many great minds, it was

10 By profession, Fermat was a lawyer and a member of the supreme court in Toulouse. He did mathematics only as a hobby. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 49

Figure 3.18 not until the year 1994 that this so-called Fermat’s Last Theorem was finally proved by Andrew Wiles. On the modest side, as always, we will here reduce the problem to n an odd prime, and give some fragmentary historical notes. We set n 4 and follow Fermat, who actually jotted down a sketch proof for this case, and Euler, who gave the first complete proof for this in 1747. More generally, we claim that there is no all-positive integral solution of the equation a4 + b4 c2, a,b,c ∈ N. We employ here what is known as “Fermat’s method of infinite descent.” It starts with an all-positive solution a0,b0,c0, that we assume to exist and creates another all-positive solution a, b, c with cs, and s, t are relatively prime and of different parity. We claim that t is odd and s is even. ¬ If t is even and s is odd, then Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

50 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

2 2 t −s is a multiple of 4 minus 1. On the other hand, b0 is odd, so that b2 ¬ 0 is a multiple of 4 plus 1. These cannot happen simultaneously! We can now write s 2r. Substituting this into the expression a2 for 0, we obtain a 2 0 rt. 2 This reminds us of an analogous equation for Pythagorean triples! Since r and t are relatively prime, it follows in exactly the same way that r and t are pure squares: t c2 and r d2. Next we write b2 s2 + b2 t2 the expression for 0 in the Pythagorean form 0 and apply the description of Pythagorean triples again to get 2 2 2 2 s 2uv, b0 u − v ,t u + v , where u>v, and u, v are relatively prime and of different parity. The equation uv s/2 r d2 tells us that u and v are pure squares: u a2 and v b2. We ﬁnally have c2 t u2 + v2 a4 + b4, so that (a,b,c)is another solution! Comparing the values of c and c0,weget 2 2 2 2 c ≤ c t ≤ t

In particular, a4 + b4 c4 does not have any all-positive solutions. Now look at the general case an + bn cn.Ifn is divisible by 4, say, n 4k, then we can rewrite this as (ak)4 + (bk)4 (ck)4, and, by what we have just proved, there is no positive solution in this case either. Assume now that n is not divisible by 4. Since n ≥ 3, this implies that n is divisible by an odd prime, say p, and we have n kp. We can then write the original equation as (ak)p + (bk)p (ck)p and Fermat’s Last Theorem will be proved if we show that, for p an odd prime, no all-positive solutions exist for ap + bp cp, Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

3. Rationality, Elliptic Curves, and Fermat’s Last Theorem 51 or equivalently, there are no rational points (with positive coordinates) on the Fermat curve deﬁned by the equation

xp + yp 1.

Now, some history. The first case, p 3, although it seems to have attracted attention even before a.d. 1000, was settled by Euler in 1770 with a gap filled by Legendre. Around 1825, Legendre and Dirichlet independently settled the next case, p 5. The next date is 1839, when Lame´ succesfully completed a proof for p 7. With the proofs getting more and more complex, it became clearer and clearer that a good way to attack the problem was to consider numbers that are more general than integers. A good class of numbers turned out to be those that are roots of polynomial equations with integral or rational coefficients. We will investigate these in the next section. Kummer went further and, introducing the so-called “ideal numbers,” managed to prove Fermat’s Last Theorem for a large class of “regular” primes. Be- fore Wiles’ recent proof, one has to mention a result of Faltings in 1983 that implies that there may be only finitely many solutions (a,b,c)for a given odd prime p. A brief account on the final phase in proving Fermat’s Last Theorem is as follows. ♠ The work of Hel- legouarch between 1970 and 1975 revealed intricate connections between the Fermat curve and elliptic curves. This led to a sugges- tion made by Serre that the well-developed theory of elliptic curves should be exploited to prove results on Fermat’s Last Theorem. In 1985, Frey pointed out that the elliptic curve y2 x(x + ap)(x − bp), where ap + bp cp, a, b, c ∈ N, is very unlikely to exist due to its strange properties. Working on the so-called Taniyama–Shimura conjecture, in 1986, Ribet proved that the Frey curve above is not modular,11 that is, it cannot be parametrized by “modular functions” (in a similar way as the unit circle given by the Pythagorean equation x2 + y2 1 can be parametrized by sine and cosine). Finally,

11 For a good expository article, see R. Rubin and A. Silverberg, “A Report on Wiles’ Cambridge Lectures,” Bulletin of the AMS, 31, 1 (1994) 15–38. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

52 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

in a technical paper, Wiles showed that elliptic curves of the form y2 x(x − r)(x − s) can be parametrized by modular functions provided that r and s are relatively prime integers such that rs(r − s) is divisible by 16. This, applied to the Frey curve with r −ap and s bp ﬁnally gives a contradiction since rs(r − s) apbpcp is certainly divisible by 16 for p ≥ 5 (as one of the numbers a, b, c must be even).

Problems

1. Following Euclid, prove the Pythagorean theorem by working out the areas of triangles in Figure 3.19. 2. Show that the radius of the inscribed circle of a right triangle with integral side lengths is an integer. 3. Use the general form of Pythagorean triples to prove that 12|ab and 60|abc for any Pythagorean triple (a,b,c). 4. Show that the only Pythagorean triple that involves consecutive numbers is (3, 4, 5). 5. Find all integral solutions a, b ∈ Z of the equation a2b + (a + 1)2b (a + 2)2b.

Figure 3.19 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 53

6. Find all right triangles with integral side lengths such that the area is equal to the perimeter. 7. Show that a2 + b2 c3 has infinitely many solutions. 8. Use the chord-method to find all rational points on the curves: (a) y2 x3 + 2x2; (b) y2 x3 − 3x − 2. 9. Consider Bachet’s curve y2 x3 + c, c ∈ Z. Show that if (x, y), y 0, is a rational point, then x4 − cx −x6 − cx3 + c2 8 , 20 8 4y2 8y3 is also a rational point on this curve. ♥ Use calculus to verify that this is the second intersection of the tangent line to the curve at (x, y). √ 10. Find all rational points on the circle with center at the origin and radius 2. 11. Prove that on an elliptic curve in Weierstrass form y2 P(x), the points ( O) of order 2 are the intersection points of the curve with the first axis. √ 12. (a) Devise a proof of irrationality of 2 using Fermat’s method√ of infinite descent. (b) Demonstrate (a) by paper folding: Assume that 2 a/b, a, b ∈ N, consider a square paper of side length b and diagonal length a, and fold a corner of the square along the angular bisector of a side and an adjacent diagonal. 13. Use a calculator to work out the first two iterates of the duplication formula for the Bachet equation y2 x3 − 2, starting from (3, 5). 14. Does there exist a right triangle with integral side lengths whose area is 78? 15. Use the result of Gauss in Section 2 on the rational roots of a polynomial with integer coefficients to describe the rational points on the graph of the polynomial. 2 3 16. ♠ Let C be the cubic cusp given by y x . (a) Show that the set Cns(Q ) of all nonsingular rational points (“ns” stands for nonsingular) forms a group under addition given by the chord method (the identity is placed at the vertical ideal point). (b) Verify that the map φ : Cns(Q ) → Q defined by φ(x, y) x/y and φ(O) 0 is an isomorphism. (Since the additive group of Q is not finitely generated, this shows that Mordell’s theorem does not extend to singular curves!) 17. Derive the analytical formulas for adding points on an elliptic curve in Weier- strass form in the case where the identity is placed at the vertical ideal point. 18. Placing the identity O at the vertical ideal point, show that the integer point A (2, 3) has order 6 in the group of rational points of the Bachet curve given by y2 x3 + 1. (Hint: Rewrite A ∗ A A ∗ A ∗ A ∗ A. Note that this is the largest finite order a rational point can have on a Bachet curve.) Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

54 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem

19. Show that the cubic curve given by x3 + y3 c can be transformed into a Weierstrass form by the rational transformation a + y a − y (x, y) → , . bx bx

Choose a 36c and b 6 to obtain the Bachet equation y2 x3 − 432c2. (Notice that c 1 gives the birational equivalence of the Fermat curve in degree 3 and the Bachet curve given by y2 x3 − 432. Since the former has (1, 0) and (0, 1) as its only rational points, it follows that (away from O, the vertical ideal point) the Bachet curve also has only two rational points. Can you ﬁnd them? With O, these form a cyclic group of order 3.) 20. Show that if 1 were a congruent number, then the equation a4 − c4 b2 would have an all-positive integral solution with b odd, and a, c relatively prime. (Hint: If 1 were congruent, then u4 − 1 v2 would have a rational solution. Substitute u a/c and v b/d, where a, c and b, d are relatively prime. Note that a Fermat’s method of inﬁnite descent, resembling the one in the text, can be devised to show that a4 − c4 b2 has no all-positive integral solutions. It thus follows that 1 is not a congruent number.) 21. Use Tunnell’s theorem to show that 5, 6, 7 are congruent numbers.

Web Sites

1. www-groups.dcs.st-and.ac.uk/∼history/HistTopics/ Fermat’s last theorem.html 2. www.math.niu.edu/∼rusin/papers/known-math/elliptic.crv/ Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Algebraic or . . . . . Transcendental? . 4 . SECTION .

♣ We managed to split the real numbers into two disjoint subsets: the rationals and irrationals. Is there a further√ split of the irrationals? For example, which is more subtle: 2ore? For the answer, we go back to Q and make the following observation: If x ∈ Q , then, writing x a/b, a, b ∈ Z, we see that x is the root of the linear equation

a − bx 0

√with integral coefﬁcients. Raising the degree by one, we see that 2 has the same property; i.e., it is a root of the quadratic equation

x2 − 2 0, again with integral coefﬁcients. We are now motivated to introduce the following deﬁnition: A real number r is algebraic if it is a root of a polynomial equation

n c0 + c1x + ...+ cnx 0,cn 0 with integral (or what is the same, rational; see Problem 1) coef- ﬁcients. The least degree n is called the degree of r. A number is called transcendental if it is not algebraic. 55 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

56 4. Algebraic or Transcendental?

We see immediately√ that the degree 1 algebraic numbers are the rationals and that 2 is an algebraic number of degree 2. What about e and π? It turns out that they are both transcendental, but the proof (especially for π) is not easy. A proof of transcendentality for e was first given by Hermite in 1873 and simplified considerably by Hilbert in 1902. Transcendentality of π was first proved by Linde- mann.1 Toappreciate these revolutionary results, one may note the scepticism that surrounded transcendentality in those days (especially coming from the constructivists). As Kronecker, the leading contemporary to Lindemann, noted: “Of what use is your beautiful research on π? Why study such problems, since there are no irrational numbers at all?” ♥ From the point of view of abstract algebra, Q is a subfield of R. Are there any fields between Q and R? The answer is certainly yes; just pick an irrational number r and consider the smallest subfield of R that contains both Q and r. This subfield is denoted by Q (r). The structure of Q (r) depends on whether r is algebraic or transcendental. √ ♣ To see what happens√ when r is algebraic, consider r √2. Since all even powers of 2areinN ⊂ Q , an element of Q ( 2) is of the form √ a1 + b1 2 √ ,a1,a2,b1,b2 ∈ Q , a2 + b2 2

with the assumption that a2 and b2 do not vanish simultaneously. Rationalizing the denominator now gives √ √ √ a + b a − b a a − b b + (a b − a b ) 1 1√2 · 2 2√2 1 2 2 1 2 2 1 1 2 2 a2 − b2 a2 + b2 2 a2 − b2 2 2 2 2 a a − 2b b a b − a b √ 1 2 1 2 + 2 1 1 2 2. a2 − b2 a2 − b2 2 2 2 2 2 2

1For an interesting account, see F. Klein, Famous Problems of Elementary Geometry, Chelsea, New York, 1955. For a recent treatment, see A. Jones, S. Morris, and K. Pearson, Abstract Algebra and Famous Impossibilities, Springer, 1991. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

4. Algebraic or Transcendental? 57 √ The fractions are rational, so we conclude that Q ( 2) consists of numbers of the form √ a + b 2,a,b∈ Q . √ In particular, Q ( 2) is a vector space√ of dimension 2 over Q (with respect to ordinary√ addition in Q ( 2) ⊂ R and multiplication of elements in Q ( 2) by Q ). The generalization is clear. Given an algebraic number r of degree n over Q , the ﬁeld Q (r) is a vector space over Q of dimension n. Example −1 For n ∈ N, deﬁne Tn,Un :[−1, 1] → R by Tn(x) cos(n cos (x)) −1 −1 and Un(x) sin((n + 1) cos (x))/ sin(cos (x)), x ∈ [−1, 1]. By the addition formulas, we have cos((n + 1)α) cos(nα) cos(α) − sin(nα) sin(α), sin((n + 2)α) sin((n + 1)α) cos(α) + cos((n + 1)α) sin(α). For α cos−1(x), these can be rewritten as 2 Tn+1(x) xTn(x) − (1 − x )Un−1(x),

Un+1(x) xUn(x) + Tn+1(x).

Since T1(x) x and U0(x) 1, these recurrence relations imply that Tn and Un are polynomials. We obtain that cos(π/n) is algebraic, since it is a root of the polynomial Tn + 1. (What about sin(π/n)?)

If r is transcendental, then Q (r) is isomorphic to the field of all rational functions a + a r +···+a rn 0 1 n ,a b , m n 0 m b0 + b1r +···+bmr with integral coefficients. In this case, we have every reason to call r a variable over Q . ♦ We finish this section by exhibiting infinitely many transcendental numbers. Let a ≥ 2 be an integer. We claim that ∞ r 1 aj! j 1 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

58 4. Algebraic or Transcendental?

is transcendental. To show convergence of the inﬁnite series, we ﬁrst note that

1 ≤ 1 , aj! 2j

with sharp inequality for j ≥ 2. Thus

∞ ∞ 1 1 1 < − 1 1, aj! j − ( / ) j 1 j 1 2 1 1 2

and so the series deﬁning r converges, and the sum gives a real number r ∈ (0, 1). Now for transcendentality: ¬ Assume that r is the root of a polynomial

n P(x) c0 + c1x +···+cnx ,c0,...,cn ∈ Z,cn 0.

For k ∈ N, let

k 1 rk aj! j 1

be the kth partial sum. We now apply the Mean Value Theorem of calculus for P on [rk,r] and obtain θk ∈ [rk,r] such that the slope of the line through (rk,P(rk)) and (r, P(r)) (r, 0) is equal to the slope of the tangent to the graph of P at θk:

P(r) − P(rk) P (θk) r − rk

(Figure 4.1). Taking absolute values, we rewrite this as

|P(rk)| |P(r) − P(rk)| |r − rk|·|P (θk)|.

l We now estimate each term as follows: For l 0,...,n, clrk is a rational number with denominator al·k!. Thus,

n n·k! |P(rk)| |c0 + c1rk +···+cnrk |≥1/a . Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

4. Algebraic or Transcendental? 59

θ rk k r

Figure 4.1

Second, we have ∞ 1 |r − rk| aj! j k+1 1 1 1 1 + + +··· a(k+1)! a(k+2)!−(k+1)! a(k+3)!−(k+1)!

< 1 1 , a(k+1)! 1 − (1/a) where we applied the geometric series formula (after comparison). Finally, taking the derivative of P: n−1 |P (θk)| |c1 + 2c2θk +···+ncnθk |

≤|c1|+2|c2|+···+n|cn| c, c ∈ N, where we used 0 ≤ θk ≤ 1. Putting all these together, we arrive at c 1 ≤ 1 . an·k! a(k+1)! 1 − (1/a) For k large, this is impossible since a(k+1)! grows faster than an·k!. ¬ For a 10, we obtain transcendentality of the number ∞ 1 0.110001000000000000000001000 .... j! j 1 10 This goes back to Liouville in 1844. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

60 4. Algebraic or Transcendental?

Remark 1. A deep theorem of Gelfond and Schneider, proved in 1935, asserts s that if r 0, 1 is algebraic and s is not rational, then r is transcen-√ dental. (r, s can both be complex; see Section 5.) For example, 2 2 and eπ ( i−2i; see Section 15) are transcendental. In 1966, Alan Baker2 found effective close approximations for sums of natural logarithms, and proved transcendence results for sums of num- a b a, b bers of the form ln , where√ √ are algebraic. In particular, he reproved transcendence of 2 2 and transcendence of the Gregory −1 numbers ta/b tan (b/a), a, b ∈ N.(ta/b is the imaginary part of the complex logarithm of the Gaussian integer a + bi ∈ Z[i].) Fi- nally, going back to geometry, we note that according to a result of Schneider in 1949, the perimeter of an ellipse with algebraic semimajor axes is transcendental.

Remark 2. We saw in Section 2 that the only rational point on the graph of the exponential function is (0, 1). As Lindemann proved, with the exception of (0, 1), the graph actually avoids all points with algebraic coordinates.

Problems

1. Show that if a real number is the root of a polynomial with rational coefﬁcients, then it is also the root of a polynomial with integer coefﬁcients. √ √ 2. Show that the numbers a − b, a, b ∈ N, are algebraic. What is the degree? √ 3. Let c be a positive real number. Prove that if c is algebraic, then so is c.

4. Verify that if c1 and c2 are algebraic, then so is c1 + c2. √ √ 5. ♠ Show that the ﬁelds Q ( 2) and Q ( 3) are not isomorphic. 6. Prove the divison algorithm for polynomials: Given polynomials P, S there exist polynomials Q, R such that P QS + R, where R is zero or has degree less than the degree of S. Moreover, Q and R are unique. In addition, show that if P and S have rational coefﬁcients, then so do Q and R.

2See “Linear Forms in the Logarithms of Algebraic Numbers I–II–III–IV,” Mathematica, 13 (1966) 204–216; 14 (1967) 102–107; 14 (1967) 220–228; 15 (1968) 204–216. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 61

p P 7. Let be a prime. Assume√ that a polynomial with rational coefﬁcients√ has a + b p a, b ∈ a − b p a root of the form , where √ Q . Show√ that is also a root. (Hint: Divide P by (x − (a + b p))(x − (a − b p)), and consider the remainder.) 8. ♠ Generalize Problem 7 to polynomials over a ﬁeld and a quadratic extension. 6 3 9. Use a calculator to show that the polynomial x − 7.5x − 19x + 2.1 evaluated ∞ / j! on the Liouville’s number j 1 1 10 vanishes to 8-decimal precision. 10. ♦ Let m 1 be positive. Show that the equation ex mx + 1

has a unique nonzero solution x0. In addition, show that x0 is irrational if m is rational, and x0 is transcendental if m is algebraic. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Complex Arithmetic . . . 5 . SECTION .

♣ We now turn the setting around and ask the following: Given a polynomial equation

n c0 + c1x +···+cnx 0,cn 0,c0,c1,...,cn ∈ R,

what can be said about the roots? This is one of the oldest problems of algebra. The factor theorem says that there are at most n roots. It also says that it is enough to prove the existence of one root r, since after dividing by the root factor (x − r), the quotient has degree n − 1, to which the existence result can again be applied. If we pretend that we do not know the FTA (fundamental theorem of algebra), then the existence of roots seems to carry some inherent difﬁculty. For example, the quadratic equation

x2 + 1 0

is unsolvable in R since the left-hand side is always ≥ 1forx real. This, of course, we knew all along. In general, the solutions of the quadratic polynomial equation

ax2 + bx + c 0 62 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

5. Complex Arithmetic 63 are given by the quadratic formula √ −b ± b2 − ac r ,r 4 , 1 2 2a with r1 corresponding to the positive and r2 to the negative sign in front of the square root. This is an easy exercise in completing the square. (To show this in the Greek way, we first divide through by a and obtain x2 + px + q 0, where p b/a and q c/a. Consider the expression x2 + px as the area of a square with side length x plus the areas of two rectangles of side lengths p/2 and x. When the rectangles are joined with the square along two adjacent sides of the square, they form a larger square of side length x + p/2 with a small square of side length p/2 missing. Inserting the missing small square completes the big square, and we obtain x2 + px + (p/2)2 (x + p/2)2. The quadratic formula follows, since x2 + px + q (x + p/2)2 − (p/2)2 + q, so that the roots are −p/2 ± (p/2)2 − q.) Now, depending on whether b2 − 4ac is positive, zero, or negative, we have two, one, or no roots among the reals. Thus, it is clear that real numbers are insufficient and that we have to introduce new numbers. These should form a field if we want to do ‘decent’ mathematics. Among the new numbers, there must be one that is a root of x2 + 1 0, as above. Thus, we introduce the complex unit i with the property i2 + 1 0. The complex unit i is sometimes denoted by (−1) for obvious reasons.√ We will use only the√ former√ to prevent silly errors such as 1 1 (−1)(−1) −1 −1 −1. For degree 2 polynomials, the problem is now solved. Indeed, for b2 − 4ac < 0, the quadratic formula can be written as √ b −(4ac − b2) b ac − b2 r ,r − ± − ± 4 i. 1 2 2a 2a 2a 2a (Notice that the first equality is somewhat heuristic; nevertheless the final expression does satisfy the quadratic equation.) Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

64 5. Complex Arithmetic

We now consider the ﬁeld R(i) (an algebraic extension of R of degree 2) consisting of all elements of the form

a1 + b1i ,a1,a2,b1,b2 ∈ R, a2 + b2i

with the assumption that a2 and b2 do not vanish simultaneously. (The reason for considering only linear fractions is clear; all even powers of i are real.) We now “rationalize the denominator”: a + b i a + b i a − b i (a a + b b ) (a b − a b ) 1 1 1 1 · 2 2 1 2 1 2 + 2 1 1 2 i. a + b i a + b i a − b i a2 + b2 a2 + b2 2 2 2 2 2 2 2 2 2 2 The fractions are real, so we conclude that R(i) consists of numbers of the form z a + bi, a, b ∈ R. For notational convenience, we write C R(i) and call it the complex number field. The typical element z above is called a complex number. a !(z) is the real part of z and b "(z) is the imaginary part of z. The complex number z is real iff "(z) 0 and purely imaginary iff !(z) 0. Notice finally that rationalizing the denominator as above gives a formula for complex division! In√ view of the analogy between the algebraic field extensions Q ( 2) in Section 4 and R(i) above, it is perhaps appropriate to remember Titchmarsh’s words: √ There are certainly many people who regard 2 as something perfectly obvious, but jib at (−1). This is because they think they can visualize the former as something in physical space, but not the latter. Actually (−1) is a much simpler concept. Summarizing, every quadratic polynomial equation has exactly two roots among the complex numbers. If b2 − 4ac 0, we say that we also have two roots that happen to coincide. (The elegant phrase “root with multiplicity 2” is not without reason; think again of the corresponding root factors!) We now pin our hopes on C for solving cubic, quartic, quintic, etc. polynomial equations. Our doubts are not completely unfounded if we go back to Roman times when only rational numbers were accepted, and pretend that we want to invent reals by considering Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

5. Complex Arithmetic 65 √ Q ( 2) (and alike), which is deﬁnitely not the whole of R. Fortu- nately this is not the case, and we can stop pretending that we did not know the

Fundamental Theorem of Algebra. Every polynomial equation n c0 + c1z +···+cnz 0,cn 0, with complex coefﬁcients c0,c1,...,cn ∈ C, has a complex root.

Having been stated as early as 1629 by Albert Girard, the FTA was ﬁrst proved by Gauss in 1799. Earlier attempts were made by D’Alembert (1746), Euler (1749), and Lagrange (1772). Many elementary proofs exist. A typical proof uses Liouville’s theorem in complex analysis which, in turn, is based on a trivial estimate of the Cauchy formula. (By the way, have you noticed that almost everything in one-variable complex calculus goes back to the Cauchy formula?) To do something out of the ordinary, we prove the FTA by a “differential topological” argument.1 Note that four additional proofs are outlined in Problems at the end of Section 8. Before proving the FTA, we explore some geometric features of complex arithmetic. To deﬁne a complex number z a + bi, we need to specify its real and imaginary parts a and b. Put together, they form a point (a, b) on the Cartesian plane or, similarly, they form a plane vector. This representation of a complex number as a plane vector is called the Argand diagram. This geometric representation of complex numbers is named after Jean Robert Argand, 1806, although nine years earlier it had been announced by the Norwegian cartographer Caspar Wessell before the Danish Academy of Sciences. Youmay say, “Well, in Euclidean plane geometry we talked about points on the plane. In calculus, we were told that a vector drawn from the origin is essentially the same thing as a point. Now we are saying that a planar point can also be thought of as a complex

1See J. Milnor, Topology from the Differentiable Viewpoint, The University Press of Virginia, 1990. For a recent comprehensive study of the FTA and its relation to various mathematical disciplines, see B. Fine and G. Rosenberger, The Fundamental Theorem of Algebra, Springer, 1997. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

66 5. Complex Arithmetic

number, an element of an abstract field. Why can we not decide on just one of these?” The answer is that all three are conceptually different, but are represented by the same mathematical model. As you go on, these kinds of identifications become more and more common. Thus, there is a one-to-one correspondence between C and the plane R2; we are thereby entitled to call this the Gauss plane or complex plane. The field operations in C carry over to operations on planar vectors. Addition gives nothing new; z1 a1 + b1i and z2 a2 + b2i add up to

z1 + z2 (a1 + a2) + (b1 + b2)i, and we see that complex addition corresponds to the usual “parallelogram rule” for planar vectors. The product

z1z2 (a1a2 − b1b2) + (a1b2 + a2b1)i

is more subtle and needs to be analyzed. To do this, we rewrite z1 and z2 in polar coordinates:

z1 r1(cos θ1 + i sin θ1),

z2 r2(cos θ2 + i sin θ2) and compute

z1z2 r1r2(cos θ1 cos θ2 − sin θ1 sin θ2)

+ r1r2(cos θ1 sin θ2 + sin θ1 cos θ2)i

r1r2(cos(θ1 + θ2) + i sin(θ1 + θ2)), where we used the addition formulae for sine and cosine. This tells us two things: First, introducing the absolute value, or modulus, of a complex number z a + bi r(cos θ + i sin θ) as the length of the corresponding vector |z| a2 + b2 r, we have

|z1z2| |z1|·|z2|. Second, introducing the multiple-valued argument arg z {θ + 2kπ | k ∈ Z}, Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

5. Complex Arithmetic 67 we also have

arg (z1z2) arg z1 + arg z2

(as sets!). The geometry behind this is the following: Consider arg : C → R as a multiple-valued function on C R2. Its graph is a surface in R3 that resembles an inﬁnite spiral staircase; multiplying z1 by z2 has the effect of walking up and down the staircase (see Color Plate 2b). In particular, consider complex numbers of modulus one:

z(θ) cos θ + i sin θ, θ ∈ R.

They ﬁll the unit circle

S1 {z ∈ C ||z| 1}.

Since

z(θ1)z(θ2) z(θ1 + θ2), θ1,θ2 ∈ R,

S1 is a multiplicative subgroup of C# C −{0}. ♥ In fact, as the identity shows, θ → z(θ), θ ∈ R, is a homomorphism of the additive group R onto the multiplicative group S1 with kernel 2πZ. Note also that, using the complex exponential function, z(θ) eiθ by the Euler formula (proved in Section 15). Since we are using only the multiplicative property of z(θ) above, we can postpone the introduction of complex exponentials. ♣ Fix n ∈ N. The complex numbers

z(2kπ/n), k 0, 1,...,n− 1, are (for n ≥ 3) the vertices of a regular n-sided polygon inscribed in S1 (with a vertex on the positive first (or real) axis). By the identity above, for fixed n these vertices form a multiplicative subgroup 1 # of S (and thereby of C ) that is isomorphic with Zn, the group of congruence classes of integers modulo n. This is a convenient way to view Zn, since we can use the field operations in C to discover some of the geometric features of this configuration. Here is one: Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

68 5. Complex Arithmetic

Proposition 3. For n ≥ 2, we have n−1 kπ z 2 . n 0 k 0

Remark. Geometrically, the sum of vectors from the centroid to the vertices of a regular polygon is zero.

Proof. This is clear for n even, but we give a proof that is independent of the parity of n. First, by complex multiplication, kπ π k z 2 z 2 , n n so that n− 1 2kπ z + ω + ω2 +···+ωn−1, n 1 k 0 where ω z(2π/n). This looks very familiar! In fact, switching from real to complex, the proof of Proposition 2 in Section 2 gives 1 − zn 1 + z + z2 +···+zn−1 ,z 1,z∈ C. 1 − z Substituting z ω and noting that ωn z(2nπ/n) z(2π) 1, the proposition follows.

The ﬁrst idea used in the proof gives another interpretation of the numbers z(2kπ/n), k 0,...,n− 1. In fact, raising them to the nth power, we get kπ n knπ z 2 z 2 z( πk) , n n 2 1 so that they all satisfy the equation zn − 1 0. By the complex version of the factor theorem, there are at most n roots; so we got them all! We call z(2kπ/n), k 0,...,n− 1, the Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

5. Complex Arithmetic 69 nth roots of unity. The number ω z(2π/n) is called a primitive nth root of unity. The powers ωk z(2kπ/n), k 0,...,n− 1, give all nth roots of unity. We now begin to believe that complex numbers should sufﬁce, and that after all, the FTA ought to be true! The nth roots of a complex number z can be written down in a convenient way using the polar form z |z|·z(θ), θ ∈ arg z. Indeed, we have √ n z n |z| n z(θ + 2kπ) n |z|z(θ/n)ωk,k∈ Z. For z 0, these complex numbers are distinct for k 0,...,n− 1. Notice that they have the same absolute value and are equally spaced. Geometrically, they are the vertices√ of a regular n-sided polygon inscribed in a circle with radius n |z| and center at the origin.

Remark. If p ≥ 3 is odd, then the solutions of the equation zp + 1 0 are the negatives of the pth roots of unity z −z(2kπ/p) −z(2π/p)k, k 0,...,p− 1. Setting ω z(2π/p), the factor theorem says that zp + 1 (z + 1)(z + ω) ···(z + ωp−1). Letting z a/b, a, b ∈ Z, we obtain the identity ap + bp (a + b)(a + ωb) ···(a + ωp−1b).

Going back to Fermat’s Last Theorem, we see that if there is an all-positive solution to ap + bp cp, then cp (a + b)(a + ωb) ···(a + ωp−1b) must hold. ♥ The right-hand side is a factorization of cp with factors not quite in the ring Z but in the ring Z[ω] obtained from Z by “adjoining” ω. If the factors have no common divisors in this extended ring, then, based on the analogy with the unique factorization in Z, we would think that each factor must be a pth power. Looking Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

70 5. Complex Arithmetic

back to Pythagorean triples, we see that this would be an essential step toward a resolution of Fermat’s Last Theorem. This idea goes back to Kummer.2 Unfortunately, in Z[ω], unique factorization does not hold for all p (in fact, the ﬁrst prime when it fails is p 23 as was recognized by Cauchy), so he had to adjoin further “ideal numbers” to this ring. In this way, he managed to prove Fermat’s Last Theorem for a large class of primes p.

Toillustrate how far-reaching the validity of unique factorization in Z[ω] is, we now consider the case n 4(ω i) and prove that unique factorization in the square lattice Z[i] of Gaussian integers (cf. Section 2) implies that any prime p of the form 4m+1 is a sum of squares of two integers. This is a result of Fermat already discussed at the end of Section 1. Let p 4m + 1, m ∈ N, be a prime in Z.By Wilson’s criterion for primality (usually proved in an introductory course in number theory), p must divide (p − 1)! + 1. We write the latter as (p − 1)! + 1 (4m)! + 1 1 · 2 ···(2m − 1)(2m)(2m + 1)(2m + 2) ···(4m − 1)(4m) + 1. Modulo p this is equal to

1·2 ···(2m−1)(2m)(−2m)(−2m+1) ···(−2)(−1)+1 (2m)!2 +1.

This number lives in Z, but factors in Z[i]as

(2m)!2 + 1 ((2m)! + i)((2m)! − i).

Since p divides the left-hand side in Z, it also divides the right- hand side in Z[i]. Clearly, p cannot divide the factors. Thus, by a mild generalization of Euclid’s characterization of primes (Book IX of the Elements), p cannot be a prime in Z[i]! Hence p must have proper divisors. A proper divisor must have the form a + bi, where b 0, since p is a prime in Z. Since p is real, the conjugate a − bi

2Emil Grosswald’s book Topics from the Theory of Numbers (Macmillan, New York, 1966) gives an excellent account of Kummer’s work. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 71 also divides p. We obtain that (a + bi)(a − bi) a2 + b2 divides p. Since a2 + b2 is real and p is a prime in Z, we arrive at p a2 + b2.

Problems

2 1. Show that the roots r1,r2 of a quadratic polynomial ax +bx+c can be obtained in the following geometric way:3 Consider the points O (0, 0), A (1, 0), B (1, −b/a), and C (1 − c/a, −b/a). Let the circle with diameter OC intersect the line through A and B at points P and Q . Then r1 and r2 are the (signed) lengths of the segments AP and AQ . Is there a generalization of this construction to cubic polynomials? √ 2. Find all values of 3 −i.

3. Let z1,z2,z3,z4 ∈ C with z1 z2 and z3 z4. Show that the line through z1,z2 is perpendicular to the line through z3,z4 iff !((z1 − z2)/(z3 − z4)) 0. 4. Factor the quartic polynomial z4 + 4ﬁrstoverC and then over R.

5. Show that the triangles with vertices z1,z2,z3 and w1,w2,w3 are similar iff the complex determinant

111 z z z . 1 2 3 w1 w2 w3 is equal to zero. √ 6. ♥ Show that unique factorization does not hold in Z[ 5i]. 2 2 2 7. Use the complex identity |z1| |z2| |z1z2| , z1,z2 ∈ C, to show that if a and b are both sums of squares of integers, then ab is also a sum of squares of integers. ♥ Use the results of Section 1 to prove the following theorem of Fermat: A positive integer a is the sum of two squares of integers iff in the prime factorization of a the primes 4k + 3 occur with even exponents.

3Cf. G.H. Hardy, A Course of Pure Mathematics, Cambridge University Press, 1960. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

. . . . . Quadratic, Cubic, . . . . . and Quartic . . . 6 . SECTION . Equations

†1 ♣ Beyond the pure existence of complex roots of polynomials of degree n, guaranteed by the FTA, the next question to ask concerns their constructibility in terms of radicals such as the quadratic formula for n 2. Although the quadratic case was known around 300 b.c. by the Greeks,2 it was not until Fran¸cois Viete` (1540–1603) that the quadratic formula was cast in its present form, due to its creator’s insistence on systematic use of letters to represent constants and variables. For later purposes, it is instructive to derive the quadratic formula in a somewhat nontraditional way. In what follows it will be convenient to normalize our polynomials by dividing through by the leading coefﬁcient. We will call a polynomial with leading coefﬁcient 1 monic. To emphasize that we are dealing with complex variables, we replace the real variable x by the complex variable z. In particular, we take our quadratic polynomial in

1This symbol indicates that the material in this section is more technical than the average text. 2The fact that extraction of square roots can be used to solve quadratic equations had already been recognized by the Sumerians. Furthermore, an ancient Babylonian tablet (c. 1600 b.c.) states problems that reduce to the solution of quadratic equations. 72 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

6. Quadratic, Cubic, and Quartic Equations 73 the form 2 P(z) z + pz + q (z − z1)(z − z2). Here we have already factored P into root factors. The complex roots z1,z2 are to be determined. We will allow the coefﬁcients p and q to be complex numbers. Multiplying out, we obtain

z1 + z2 −p and z1z2 q. These formulas are symmetric with respect to the interchange z1 ↔ z2. Since the discriminant 2 δ (z1 − z2) is also symmetric, it is reasonable to expect that we can express δ in terms of the coefﬁcients p and q. This is indeed the case, since 2 2 δ (z1 + z2) − 4z1z2 p − 4q. √ The two values ± δ ±(z1 − z2) can be combined with the expression for the sum of the roots above. We obtain √ 2 2z1, 2z2 −p ± δ −p ± p − 4q. The quadratic formula for P follows. Substituting p b/a and q c/a, we obtain the quadratic formula for the polynomial az2 +bz +c discussed in Section 5. The solution of the cubic equation (with no quadratic term) in terms of radicals was ﬁrst obtained by del Ferro in 1515. He passed this on to some of his students. In 1535, one of the students, Fiore, challenged Fontana (nicknamed Tartaglia), who had treated some particular cases of cubics, to a public contest of solving cubic equations. Before the contest, Fontana found the solution for general cubics, and iicted a humiliating defeat on Fiore. Bent on persuasion, Fontana told the trick to Cardano (in a poem), and allegedly swore him to secrecy that he would not reveal it to anyone. Nevertheless, Cardano (with generous references to del Ferro and Fontana) published it in his Ars Magna3 in 1545, and the formula was subsequently named after him. To derive the Ferro– Fontana–Cardano formula for the roots of a cubic polynomial, we

3For a brief account on the story, see Oystein Ore’s Foreword in the Dover edition of the Ars Magna. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

74 6. Quadratic, Cubic, and Quartic Equations

will employ a method due to Lagrange. This is somewhat tedious, but less ad hoc than the more traditional approach (cf. Problem 1). The further advantage of presenting the Lagrange method now is that it will reappear in a more subtle setting for the solution of quintics in Section 25. The ﬁrst step is the same for all methods. We reduce the general monic cubic equation z3 + az2 + bz + c 0, a,b,c ∈ C, to the special cubic equation P(z) z3 + pz + q 0 by means of the substitution z → z − a/3. (Once again, a generalization of this seemingly innocent reduction (termed the Tschirnhaus transformation later) will gain primary importance in solving quintic equations. We could also have performed this reduction for quadratic equations, but that would have been the same as completing the square, a standard way to derive the quadratic formula.) Assume now that z1,z2,z3 are the roots of P: 3 z + pz + q (z − z1)(z − z2)(z − z3). Multiplying out, we have

z1 + z2 + z3 0,z1z2 + z2z3 + z3z1 p, z1z2z3 −q.

Remark. As an application, we show that z1,z2,z3 are the vertices of an equilateral triangle iff z2 + z2 + z2 z z + z z + z z . 1 2 3 1 2 2 3 3 1 First, we notice that this equation remains unchanged when z1,z2,z3 are simultaneously subjected to the substitution z → z−d, where d is any complex number. (Geometrically, this corresponds to translation of the triangle by the translation vector −d.) Choos- ing d (z1 + z2 + z3)/3 (the centroid of the triangle), and adjusting the notation, we may thus assume that z1 + z2 + z3 0 holds. Since (z + z + z )2 z2 + z2 + z2 + (z z + z z + z z ) , 1 2 3 1 2 3 2 1 2 2 3 3 1 0 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

6. Quadratic, Cubic, and Quartic Equations 75 the stated criterion splits into two equations,

z1 + z2 + z3 0 and z1z2 + z2z3 + z3z1 0.

Now consider the monic cubic polynomial with roots z1,z2,z3. Expanding, we have 3 2 (z − z1)(z − z2)(z − z3) z + az + bz + c.

Our conditions translate into a b 0. Equivalently, z1,z2,z3 are 3 the solutions of the equation z + c 0. Since these are the three√ cubic roots of −c, they are equally spaced on the circle |z| 3 |c|. The claim follows. (For another solution to this problem, we could have used the factorization z2 + z2 + z2 − z z − z z − z z (z + z ω + z ω2)(z + z ω2 + z ω), 1 2 3 1 2 2 3 3 1 1 2 3 1 2 3 where ω z(2π/3) is a primitive third root of unity, but we preferred a less ad hoc approach.) Returning to the main line, once again we consider the discriminant 2 2 2 2 δ (zj − zl) (z1 − z2) (z2 − z3) (z3 − z1) . 1≤j

(For the discriminant of the√ general monic cubic, see Problem 2.) Let ω z(2π/3) (−1 + 3)/2 be a primitive third root of unity, and consider the so-called Lagrange substitutions 3 1 (j−1)l ξl ω zj,l 0, 1, 2. 3 j 1 Expanding the sums and using ω3 1, we obtain

3ξ0 z1 + z2 + z3, 2 3ξ1 z1 + ωz2 + ω z3, 2 3ξ2 z1 + ω z2 + ωz3. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

76 6. Quadratic, Cubic, and Quartic Equations

ξ ξ3 ξ3 Since our cubic is reduced, 0 vanishes. Moreover, 1 and 2 remain unchanged when the roots z1,z2,z3 are subjected to even (in our ξ3 case cyclic) permutations. Thus, it is reasonable to expect√ that 1 ξ3 p, q δ and 2 can be expressed as polynomials in , and , since the latter three seem to be more “elementary” and possess the same symmetries. Once again a somewhat tedious calculation gives √ q 3 √ q p 3 q 2 ξ3 − + i δ − + + , 1 2 18 2 3 2 √ q 3 √ q p 3 q 2 ξ3 − − i δ − − + . 2 2 18 2 3 2 The second formula follows from the ﬁrst because ξ ξ z2 + z2 + z2 + (ω + ω2)(z z + z z + z z ) 9 1 2 1 2 3 1 2 2 3 3 1 2 (z1 + z2 + z3) − 3(z1z2 + z2z3 + z3z1) −3p. Here we used Proposition 3 to the effect that 1 + ω + ω2 0. Summarizing, we obtain 3 2 3 q p q ξ − + + , 1 2 3 2 3 2 3 q p q ξ − − + , 2 2 3 2

where the cubic roots are chosen such that ξ1ξ2 −p/3 is satis- ﬁed. Finally, the Lagrange substitutions can be solved uniquely for z1,z2,z3 in terms of ξ1 and ξ2, since the determinant

11 1 √ ωω2 (ω2 − ω) − i 1 3 3 3 1 ω2 ω

is nonzero. The solution is 2 −(j−1)l zj ω ξl,j 1, 2, 3. l 1 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

6. Quadratic, Cubic, and Quartic Equations 77

Indeed, substituting we have 3 3 2 (j−1)l (j−1)l −(j−1)k ω zj ω ω ξk j 1 j 1 k 1   2 3  (j−1)(l−k) ω ξk 3ξl, k 1 j 1 since the sum in the last set of parentheses is 3 for k l and zero for k l. Putting everything together, we ﬁnally obtain the roots of the reduced cubic:

z1 ξ1 + ξ2, 2 z2 ω ξ1 + ωξ2, 2 z3 ωξ1 + ω ξ2, where ξ1 and ξ2 are given in terms of p and q above.

Remark. Recall that in the formulas for ξ1 and ξ2 we have to choose the values of the cube roots such that the constraint ξ1ξ2 −p/3 holds. With a ﬁxed choice of cube roots, z1,z2,z3 give the solutions to our cubic 2 equation. Notice, however, that ξl, ωξl, and ω ξl are the 3 cube roots 3 of ξl , l 1, 2, so that our three solutions can be (and usually are) written more concisely as 3 2 3 2 3 q p q 3 q p q − + + + − − + 2 3 2 2 3 2 with the understanding that we choose the cube roots appropriately. We now turn to quartic equations. Cardano’s student Ferrari (1522–1565) settled the problem of solving quartic equations; this is still included in Cardano’s Ars Magna. As in the cubic case, the solution of the general quartic equation

z4 + az3 + bz2 + cz + d 0, a,b,c,d ∈ C, Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

78 6. Quadratic, Cubic, and Quartic Equations

can be reduced to the special case P(z) z4 + pz2 + qz + r 0, by means of the substitution z → z − a/4. Assuming that z1,z2,z3,z4 are the roots of this reduced quartic, expanding the root factors, we obtain the following equations:

z1 + z2 + z3 + z4 0,z1z2 + z1z3 + z1z4 + z2z3 + z2z4 + z3z4 p,

z1z2z3 + z1z2z4 + z1z3z4 + z2z3z4 −q, z1z2z3z4 r. To follow our earlier path, we would need to work out the discriminant 2 δ (zj − zk) 1≤j

in terms of p, q, r, but this seems√ a rather unpleasant task. In- stead, based on the analogy with δ, we look for polynomials in z1,z2,z3,z4 that possess only partial symmetries when these variables are permuted. We set z∗ (z + z )(z + z ), 1 1 2 3 4 z∗ (z + z )(z + z ), 2 1 3 2 4 z∗ (z + z )(z + z ). 3 1 4 2 3 Each of these new variables is symmetric only with respect to speciﬁc permutations of z1,z2,z3,z4. The elementary symmetric z∗ + z∗ + z∗ z∗z∗ + z∗z∗ + z∗z∗ z∗z∗z∗ polynomials 1 2 3, 1 2 2 3 3 1 and 1 2 3 produced z∗,z∗,z∗ from 1 2 3, however, are symmetric with respect to all permutations, and thereby they should be expressible in terms of the coefﬁcients p, q, r! For example, expanding, we obtain z∗ + z∗ + z∗ (z z + z z + z z + z z + z z + z z ) p. 1 2 3 2 1 2 1 3 1 4 2 3 2 4 3 4 2 Similarly, we have z∗z∗ + z∗z∗ + z∗z∗ p2 − r z∗z∗z∗ −q2. 1 2 2 3 3 1 4 and 1 2 3 z∗,z∗,z∗ We conclude that 1 2 3 are the three roots of the cubic P∗(z) z3 − 2pz2 + (p2 − 4r)z + q2. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

6. Quadratic, Cubic, and Quartic Equations 79

But to find the roots of a cubic is exactly what we just accomplished! z∗,z∗,z∗ Thus, using our earlier reduction and formulas, 1 2 3 can be explicitly expressed in terms of p, q, r. Finally, to pass from the roots of P∗ to the roots of our original quartic P is now easy. Since P is reduced, we have z1 + z2 + z3 + z4 0, and the formulas z∗,z∗,z∗ defining 1 2 3 can be resolved, yielding z + z −z∗,z+ z − −z∗, 1 2 1 3 4 1 z + z −z∗,z+ z − −z∗, 1 3 2 2 4 2 z + z −z∗,z+ z − −z∗. 1 4 3 2 3 3 Once again, we have to fix the ambiguity inherent in the choice of the square roots. As computations confirm, we need −z∗ −z∗ −z∗ −q 1 2 3 to be satisfied. Finally, solving the system above for z1,z2,z3,z4, we obtain z −z∗ + −z∗ + −z∗, 2 1 1 2 3 z −z∗ − −z∗ − −z∗, 2 2 1 2 3 z − −z∗ + −z∗ − −z∗, 2 3 1 2 3 z − −z∗ − −z∗ + −z∗. 2 4 1 2 3 One final remark. Since we have z∗ − z∗ −(z − z )(z − z ), 1 2 1 4 2 3 z∗ − z∗ −(z − z )(z − z ), 1 3 1 3 2 4 z∗ − z∗ −(z − z )(z − z ), 2 3 1 2 3 4 the discriminant of P∗ is the same as the discriminant of P! From the explicit formula of the discriminant for cubics (cf. Problem 2), we obtain that the discriminant of P is δ 16p4r − 4p3q2 − 128p2r2 + 144pq2r − 27q4 + 256r3. The cubic P∗ that helped to solve the quartic equation is called the resolvent cubic. For the future discussion of quintics in Section 25 all we need to remember is that a resolvent is a polynomial Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

80 6. Quadratic, Cubic, and Quartic Equations

whose roots are prescribed functions of the roots of the original polynomial. The solvability of degree-five polynomial equations eluded mathematicians for nearly 300 years, until Abel showed the impossibility of a radical formula in 1824. (Eleven years earlier Ruffini attempted to prove the impossibility but the proof contained several gaps.) The final phase was completed by Galois (and published in 1846 posthumously, 14 years after he was killed in a duel), who gave a group-theoretical criterion as to which equations of a given degree have solutions in terms of radicals.

Problems

1. Find the Ferro–Fontana–Cardano formula for a root of a general cubic polynomial z3 + az2 + bz + c

following these steps: (a) Make the substitution z → z − a/3 and verify that in terms of z, the polynomial has the form P(z) z3 + pz + q.

(b) Make the substitution z u − v and show that the polynomial reduces to P(u − v) u3 − v3 + q − (3uv − p)(u − v). (c) Set 3uv − p 0 and u3 − v3 + q 0, solve the ﬁrst equation for v, substitute it into the second equation, and arrive at 33u6 + 33u3q − p3 0. (d) Solve this for w u3 using the quadratic formula and obtain for z u −v, q q 2 p 3 q q 2 p 3 u 3 − + + ,v 3 + + + . 2 2 3 2 2 3

(Note also that the single substitution4 z p/(3v) − v reduces the cubic P to a quadratic polynomial in v3.)

4This was suggested by Viete` and appeared in print posthumously in 1615 in De aequationum recognitione et emendatione. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 81

2. Derive the formula δ a2b2 − 4b3 − 4a3c − 27c2 + 18abc for the discriminant of the general cubic z3 + az2 + bz + c. 3. Show that the roots of a reduced cubic with real coefficients are all real iff δ ≥ 0. Assuming δ>0, derive a formula√ for the roots that involves only real 5 ξ3 −q/ ± i δ/ numbers. (Hint: Write 1,2 2 3 18 in polar form.) 4. In this (admittedly long) problem we deal with geometric constructions. We start with two points in the plane. These points will be called constructible, as will any other points that can be obtained from them by making repeated use of straightedge and compass. The rules for the constructions are the following. (i) The line passing through two constructible points is constructible. (ii) A circle with center at a constructible point and passing through another constructible point is constructible. (iii) The points of intersection of constructible lines and circles are constructible. (a) Show that the line through a constructible point and perpendicular or parallel to a constructible line is constructible. (b) By marking off the distance between two constructible points on a constructible line starting at a constructible point, prove that we arrive at a constructible point. (c) Set up Cartesian coordinates such that the two initial points correspond to (0, 0) and (1, 0), and the coordinate axes are constructible. Call a number a ∈ R constructible if |a| is the distance between two constructible points. Show that a point p (a, b) is constructible iff the components a, b are constructible. (d) Use similar right triangles and the relations a :1 ab : b and 1 : a 1/a : 1 to prove that the constructible numbers form a subfield of R. As a byproduct, conclude that all rational numbers are constructible. (e) Use Problem 1 of Section 5 to prove that a quadratic polynomial ax2 + bx + c with constructible coefficients a, b, c has constructible roots r1 and r2. (f) ♥ Show that r ∈ R is constructible iff r is contained in a subfield K of R, and there is a chain of subfields

Q K0 ⊂ K1 ⊂ ··· ⊂ Kn K √ such that for each j 1,...,n, Kj Kj−1( rj), where rj ∈ Kj−1 is a positive real number (that is not a square in Kj−1). (♠ Conclude that if r ∈ R is constructible, then the degree [Q (r) : Q ] of the field extension Q (r)/Q is a power of 2. Warning: The converse is fase! In fact, there are degree-4 field extensions of Q that contain nonconstructible numbers.) (g) ♥ Show that if a cubic polynomial x3 + ax2 + bx + c with rational coefficients has no rational root, then it has no constructible root. (Hint: ¬ Assume that r is n a constructible root√ of the cubic such that the number in (f)√ is minimal. Write r s + t rn with s, t ∈ Kn−1 and prove that r˜ s − t rn is also a root (cf. Problems 7–8 of Section 4). Verify that the third root of√ the cubic 3 is in Kn−1. ¬) Derive the following results of Wantzel (1837): (i) 2isnot

5The case δ>0 is historically called “casus irreducibilis,” since the Ferro–Fontana–Cardano formula involves complex numbers, while the roots are real. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

82 6. Quadratic, Cubic, and Quartic Equations

√ constructible. (♠ Alternatively, [Q ( 3 2) : Q ] 3.) ♥ Interpret this result in view of the Delian problem:6 Construct an altar for Apollo twice as large as the existing one without changing its cubic shape. (ii) Use Problem 3 of Section 2 to show that cos 20◦ is an irrational root of the polynomial P(x) 8x3 −6x −1, thereby not constructible. Since cos 60◦ is constructible, the impossibility of the trisection of a 60◦ angle follows. This problem dates back to the fifth century b.c. in Greek geometry. (h) Show that constructibility of the regular n-sided polygon Pn is equivalent to the constructibility of the primitive nth root of unity ω z(2π/n). (Note that according to the example in Section 4, ω is a point with algebraic coordinates.) 5. ♠ (a) Construct a regular pentagon in the following algebraic way: The fifth roots of unity, other than 1 itself, satisfy the equation z4 + z3 + z2 + z + 1 0. Divide both sides of this equation by z2, rearrange the terms, and notice that the substitution w z + 1/z reduces the equation to w2 + w − 1 0. Solve this by the quadratic formula and construct the roots (cf. Problem 1 of Section 5). (In Book IV of the Elements Euclid constructed a regular pentagon. Simpler constructions were given by Ptolemy, and also by Richmond in 1893.) (b) Apply the argument in (a) to the regular 7-sided polygon. Show that the equation z7 1 reduces to the cubic equation w3 + w2 − 2w − 1 0 and w 2 cos(360◦/7) is a solution, thereby irrational. Verify that this equation has no rational roots, so that the regular 7-sided polygon is not constructible.7 6. Solve the problem of Zuane de Tonino da Coi (c. 1540): “Divide 10 into 3 parts such that they shall be in continued proportion and that the product of the first two shall be 6.” (Hint: a + b + c 10, a/b b/c, and ab 6 give a quartic equation in b.)

6For a concise treatment of Galois theory including the solutions of some problems of antiquity, see Ian Stewart’s Galois Theory, Chapman and Hall, 1973. 7 m Gauss proved that a regular n-sided polygon, n ≥ 3, is constructible iff n 2 p1 ···pk, where pj aj 22 + 1, aj 0, 1, 2,...,j 1,...,k, are distinct Fermat primes. (To be precise, Gauss did not state explicitly that the condition that the Fermat primes are distinct is necessary; this gap was ﬁlled by Wantzel in 1837.) For a fairly comprehensive account of geometric constructions, including the construction of Gauss of the regular 17-sided polygon, see Klein, Famous Problems in Elementary Geometry, Chelsea, New York, 1955. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Stereographic . . . . . Prection . 7 . SECTION .

♣ Consider the complex plane C imbedded in R3 as the plane spanned by the ﬁrst two axes. Algebraically, we view a complex number z a + bi, a, b ∈ R, as the 3-vector (a, b, 0) ∈ R3. The set of unit vectors in R3 form the 2-sphere: S2 {p ∈ R3 ||p| 1}. (S2 intersects C in the unit circle S1.) We now pretend that we are Santa Claus; we sit at the North Pole N (0, 0, 1) and look down to Earth S2. We realize that every time we see a location on the transparent Earth, we also see a unique point on C. Associating the point on Earth with the point on the complex plane C that we see simultaneously gives the stereographic projection 2 hN : S −{N}→C.

2 Geometrically, given p ∈ S , p N, hN (p) is the unique point in C such that N, p, and hN (p) are on the same line (Figure 7.1). hN is clearly invertible. Santa Claus aside, the importance of hN is clear: it allows us to chart maps of various parts of the curved globe on a ﬂat piece of paper. If we are in Brazil, we may want to use the South Pole S (0, 0, −1) instead of the North Pole. We arrive at another 83 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

84 7. Stereographic Projection

Figure 7.1

stereographic projection

2 hS : S −{S}→C.

Looking down or up we see that hN (S) hS(N) 0, the origin at C. We can thus form the composition

−1 hN ◦ hS : C −{0}→C −{0}. We claim that z −1 N ◦ h )(z) , 0 z ∈ C. S |z|2

Fix a nonzero complex number z. Since all the actions of hN and hS are happening in the plane spanned by N, S and z, we arrive at Figure 7.2. −1 The angles S S (z) and 0zS have perpendicular sides, so Eu- clid’s Elements tells us that they are equal. Thus, the right triangles

z −1 hhN S (z)

− h 1(z) S

Figure 7.2 S Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

7. Stereographic Projection 85

−1 %0N(hN ◦ hS )(z) and %0zS are similar. Again by Euclid, we have 1 |h ◦ h−1(z)| , N S |z| and this gives the required formula. We need to rewrite z → z/|z|2 in a more convenient form by introducing complex conjugation. Given a complex number z a + bi, its complex conjugate z¯ is deﬁned as z¯ a − bi.

Crossing the Gaussian Bridge we see that conjugation corresponds to rection in the real axis. As far as our algebra is concerned, we have

z1 + z2 ¯z1 +¯z2,

z1 · z2 ¯z1 ·¯z2, as can be easily veriﬁed. The real and imaginary parts of a complex number z can be written in terms of conjugation as (z +¯z) (z −¯z) !(z) and "(z) . 2 2i Moreover, zz¯ |z|2, since zz¯ (a+bi)(a−bi) a2 +b2. This Length2-Identity is important, since it ties complex multiplication to the length of the vector corresponding to z. More generally, the dot product of z1,z2 ∈ C considered as vectors in R2, is given by (z¯ z + z z¯ ) z · z 1 2 1 2 !(z z¯ ) 1 2 2 1 2 because !(z1z¯2) !((a1 + b1i)(a2 − b2i)) a1a2 + b1b2 (with obvious notations). Cross-dividing, we see that 1 z¯ ,z 0. z |z|2 This also provides a convenient formula for complex division. Going back to our stereographic projections, we obtain −1 N ◦ hS )(z) 1/z,¯ 0 z ∈ C. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

86 7. Stereographic Projection

It is rewarding to study the geometry of the stereographic projec- 2 tion more closely. Let p ∈ S , p N, q hN (p), and let l denote the line through these points. Consider the configuration of the 2 three planes VN , Vp, and C, where VN is tangent to S at N, and Vp is tangent to S2 at p. By the geometry of the sphere, the perpendicular bisector of the segment connecting N and p contains the line VN ∩ Vp. Since VN and C are parallel, the perpendicular bisector of the segment connecting p and q hN (p) contains Vp ∩ C. Figure 7.3 depicts the situation with an edge-on view of the three planes. A plane W that contains the line l (through N, p, and q) intersects Vp ∪ C in 2 lines meeting at a point v0 ∈ Vp ∩ C. The configuration consisting of the two half-lines emanating from v0 and containing p and q and the segment connecting p and q looks like a figure A laid on its side (see Figure 7.4). We call this an A-configuration and denote it by A. The segment connecting p and q is called the median of A. Since the perpendicular bisector of the median contains v0, geometrically, an A-configuration is an isosceles triangle whose two sides are extended beyond the vertices of the base. We call these the extensions of A. The upper extension is tangent to S2 at its initial point p, and the lower extension is a half-line in C emanating from q. The inner angles of the two extensions with the median at p and at q are equal. Rotating W around l, we obtain a continuous family of A-configurations whose upper extensions sweep the tangent plane Vp around. Similarly, the lower extensions centered at q sweep C around.

Vp N VN p

1 C q

Figure 7.3 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

7. Stereographic Projection 87

Figure 7.4

Now let v ∈ Vp be a tangent vector, and let w be the image of v under the (stereographic) projection from N. Then the A- conﬁguration whose upper extension contains v also contains w in its lower extension, and the terminal points of v and w are collinear with N (see Figure 7.5). 2 Consider now two vectors v1 and v2 tangent to S at p, and let w1 and w2 be their respective projections from N. The two A- conﬁgurations A1 and A2 contain v1,w1 and v2,w2. Moreover, the segment connecting p and q is the common median of A1 and A2. By bilateral symmetry with respect to the perpendicular bisector of this common median, the angle between v1 and v2 is the same as the angle between w1 and w2. We obtain that the sterographic projection is angle-preserving! We close this section by showing that the stereographic projec- 2 tion hN is also circle-preserving in the sense that if C ⊂ S is a circle

Figure 7.5 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

88 7. Stereographic Projection

that avoids the North Pole N, then S hN (C) ⊂ C is also a circle. (If C passes through the North Pole, then S is a line. Why?) There are beautiful geometric arguments to prove this.1 For a change, we give here an algebraic proof. A circle C on S2 is the intersection of a plane with S2. Using a, b, c as coordinates, p (a,b,c) ∈ S2 ⊂ R3, the equation of the plane is given by αa + βb + γc δ,

where α, β, γ, δ ∈ R are the coefﬁcients. Since C does not pass through the North Pole, we have γ δ. Using the explicit expression of the inverse of hN (Problem 1(c)), the image hN (C) is given by !(z) "(z) |z|2 − α 2 + β 2 + γ 1 δ, |z|2 + 1 |z|2 + 1 |z|2 + 1 where we used the complex variable z on C. Letting z x + iy, x, y ∈ R, we obtain 2αx + 2βy + γ(x2 + y2 − 1) δ(x2 + y2 + 1). Equivalently, (δ − γ)(x2 + y2) − 2αx − 2βy + γ + δ 0. This is the equation of a circle, since δ − γ 0. The claim follows.

Problems

1. (a) Given p (a,b,c) ∈ S2 ⊂ R3, c 1, show that a + bi hN (p) ∈ C. 1 − c (b) Verify the identity 2 hN (−p) −1/hN (p), p ∈ S . (c) Given z ∈ C, show that z |z|2 − −1 2 1 2 hN (z) , ∈ S . |z|2 + 1 |z|2 + 1

1See D. Hilbert and S. Cohn-Vossen, Geometry and Imagination, Chelsea, 1952. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Web Site 89

(Recall that C is imbedded in R3 as the plane spanned by the ﬁrst two coordinate axes.) (d) Let ω z(2π/3) be a primitive third root of unity. For what value of r>0 are the points 0, r, rω, and rω2 the stereographically projected vertices of a regular tetrahedron inscribed in S2?

2. Using the formula for hN in Problem 1 and a similar formula for hS, work out −1 hN ◦ hS explicitly.

3. Show that z1z2 ¯z1z¯2. 4. Prove that

z − w < 1 1 − zw¯ if |z| < 1 and |w| < 1.

Web Site

1. www.geom.umn.edu/∼sullivan/java/stereop/ Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

. . . . . Proof of the . . . . . Fundamental . . . 8 . SECTION . Theorem of Algebra

♣ Consider the polynomial n P(z) c0 + c1z + ...+ cnz ,cn 0, asamapP : C → C. We “pull P up to S2” by using the stereographic 2 2 2 projection hN : S −{N}→C. Analytically, we deﬁne f : S → S by −1 ◦ P ◦ h )(p), p N f(p) N N if , N, if p N.

2 f is clearly smooth on S −{N} since hN is smooth. We claim that f is smooth1 across N. To show this, we “pull f down to S2”bythe 2 stereographic projection hS : S −{S}→C and consider

−1 Q hS ◦ f ◦ hS . Notice that smoothness of f near N corresponds to smoothness of Q near zero. We now compute

−1 Q(z) S ◦ f ◦ hS )(z)

1See “Smooth Maps” in Appendix D. 90 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

8. Proof of the Fundamental Theorem of Algebra 91

−1 −1 −1 [ N ◦ hS ) ◦ P ◦ N ◦ hS )](z) −1 −1 N ◦ hS ) (P(1/z))¯

1 P(1/z)¯ zn . n n−1 c¯0z + c¯1z +···+c¯n This is a complex rational function with nonvanishing denominator at z 0. Since any rational function is smooth wherever it is defined, we conclude that Q is smooth near z 0. Thus f : S2 → S2 is smooth near N and hence everywhere. Before going any further, we work out a particular example in order to get a feel for f . Setting P(z) z2, we first describe the transformation z → z2 on the complex plane C. Consider first what happens to the unit circle S1 ⊂ C: Since S1 is parametrized by the points z(θ) cos θ + i sin θ, θ ∈ R, and z(θ)2 z(2θ), we see that S1 is mapped onto itself, and, on S1, f corresponds to doubling the angle. Topologically, S1 is stretched to twice its perimeter and then wrapped around itself twice (Figure 8.1). Going a step further, it is now clear that any circle concentric to S1 on C is mapped to another concentric circle (with the radius changing from r to r2) and, restricted to these, the transformation has the same stretch and wraparound effect. Finally, since under 2 hN the concentric circles correspond to parallels of latitude on S , we see that f : S2 → S2 wraps each parallel twice around another parallel and keeps the North and South Poles fixed. What is

Figure 8.1 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

92 8. Proof of the Fundamental Theorem of Algebra

important to us is that, apart from the poles (which we will call “singular”), each point q ∈ S2 −{N, S} has exactly two inverse images; that is, |f −1(q)| 2. (Actually, f : S2 → S2 is called a “twofold branched covering with singular points N and S.” The wraparound effect or, more precisely, the branching is shown in Color Plate 3a with f being vertical projection to a disk.) Moreover, the singular points are N and S and the latter corresponds (under hN )tothe origin in C where the derivative of P vanishes. The description for P(z) zn, n ∈ N, is similar with two replaced by n; that is, |f −1(q)| n,forq ∈ S2 −{N, S}. ♠ We now return to the main discussion with a good clue. First we have to settle the problem of singular points. The complex derivative of P is a polynomial of degree n − 1:

n−1 P (z) c1 + 2c2z +···+ncnz . 2 −1 P vanishes on, at most, n−1 points. Let σ ⊂ S denote the (f ◦hN )- image of these points plus N. Then σ consists of at most n points. We now claim that the function # : S2 − σ → N ∪{0} defined by #(q) |f −1(q)|,q∈ S2 − σ, is actually a constant. Since S2 −σ is connected, it is enough to show that # is locally constant; that is, for every q ∈ S2 − σ, there is an open neighborhood V of q in S2 −σ such that # is constant on V. Let 2 −1 q ∈ S − σ be fixed and p ∈ f (q). The derivative P of P at hN (p) is nonzero (since p/∈ σ), so that P is a local diffeomorphism2 near hN (p). Since the stereographic projection hN is a diffeomorphism, f is a local diffeomorphism at each point of f −1(q). In particular, f −1(q) cannot have any accumulation points, so it must be finite. −1 Let #(q) m with f (q) {p1,...,pm}. (Actually, m ≤ n because a degree-n polynomial can have at most n roots, but we do not need this additional fact.) For each pj, j 1,...,m, there exists an open neighborhood Vj of pj such that f |Vj : Vj → f(Vj) is a diffeomorphism. By cutting finitely many times, we may assume that the Vj’s are mutually disjoint and they all map under f diffeo- morphically onto a single V0, an open neighborhood of q. Finally,

2We are using here some basic facts in complex calculus. For a quick review, see the beginning of Section 15 and “Smooth Maps” in Appendix D. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 93

2 we delete from V0 the closed set f(S − 1 ∪···∪Vm)) to obtain 2 an open neighborhood V of q. (Notice that S − 1 ∪···∪Vm) is closed in S2; hence, it is compact. Its f -image is also compact and 2 thus closed in S .) Clearly, # is constant ( m)onV0. The claim follows. The globally constant function # cannot be identically zero, since otherwise f would map into the ﬁnite σ and the polynomial P would reduce to a constant. Thus # is at least one on S2 − σ. In particular, S2 − σ is contained in the image of f . The latter is closed in S2 so that the image must be the whole of S2. We obtain that f is onto. Thus, P must be onto. In particular, 0 ∈ C is in the image of P. This means that there exists z ∈ C with P(z) 0. This is the FTA!

Problems Fill in the details in the ﬁrst four problems, which outline additional proofs of the FTA. 1. Let n P(z) c0 + c1z +···+cnz ,cn 0, be a polynomial with complex coefﬁcients. (a) Show that the nonnegative real function z →|P(z)| attains its global minimum on C. (Let R>0 and use the triangle inequality to show that, for |z| >R, we have |c | |c | |P(z)| > |z|n |c |− n−1 −···− 0 . n R Rn

Choose R large enough so that |P| will be above its greatest lower bound m for |z| >R. Refer to closedness and boundedness of the domain and show that |P| attains m on the disk {z ||z|≤R}.) (b) Show that if z0 ∈ C is a local minimum of |P|, then z0 isazeroofP. Assume that P(z0) 0 and show that z0 is not a local minimum of |P| as follows: Consider the polynomial Q(z) P(z + z0)/P(z0) of degree n and with constant term 1 and verify that |P| has no local minimum at z0 iff |Q | has no local minimum at 0 (iff |Q | takes values < 1 near 0). Let n Q(z) b0 + b1z +···+bnz ,bn 0,b0 1.

Let k be the least positive number with bk 0. Let r>0 and use the triangle inequality again to show that for |z| r, we have

n k+1 k |Q(z)|≤|bn|r +···+|bk+1|r +|bkz + 1|. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

94 8. Proof of the Fundamental Theorem of Algebra

k k For r small enough, this is dominated by (1/2)|bk|r +|bkz + 1|. Choose z on k k the circle |z| r such that bkz + 1 −|bk|r + 1. Comparing with the above, k for this z, |Q(z)|≤1 −|bk|r /2 < 1. 2. ♠¬Let P be a nonzero polynomial that has no complex roots. Define f as in the text and observe that the image Y of f does not contain the South Pole S. (a) Use compactness of Y to show that S has an open neighborhood disjoint from Y. (b) Use the local diffeomorphism property to prove that whenever P(z) −1 0, the point f(hN (z)) is in the interior of Y. (c) Exhibit infinitely many boundary points of Y by considering the “southernmost” point in the intersection of Y with any meridian of longitude. (d) Conclude that P vanishes at infinitely many points. ¬ 3. ¬ Let P : C → C −{0} be as in Problem 2. For each r>0, consider the closed curve wr :[0, 2π] → C −{0}, wr (θ) P(rz(θ)),0 ≤ θ ≤ 2π. Let M ⊂ R3 be the graph of the multivalued function arg : C −{0}→R with projection p : M → C −{0}. (a) Show that wr can be “lifted” to a curve w˜ r :[0, 2π] → M satisfying p ◦ w˜ r wr . (Define w˜ r locally using a subdivision 0 θ0 <θ1 <...<θm 2π of [0, 2π] into sufficiently small subintervals such that arg is single-valued on each subarc wr ([θi−1,θi]), i 1,...,m.) (b) Verify that the winding number3 (1/(2π))(w˜ r (2π) − w˜ r (0)) of wr is a nonnegative integer and is independent of the choice of the lift. Note that w˜ r is unique up to translation with an integer multiple of 2π along the third axis in M ⊂ R3. (c) Prove that for r large, the winding number is n, the degree of P. (arg wr is increasing in θ ∈ [0, 2π]forr large. For this, work out the dot product (iwr ) · (∂/∂θ)wr and use the first estimate in Problem 1 to conclude that it is positive for r large.) (d) Observe that for r small, the winding number is zero. (e) Use a continuity argument to show that the winding number is independent of r. ¬ 4. Let P be a polynomial of degree n as in Problem 1. Assume first that P has real coefficients. (a) If n is odd, use calculus to show that P has a real root. (In particular, for n 3, this means that R3 is not an extension field of R, so that, unlike R2, no multiplication makes R3 a field.) (b) In general, write n 2ma, where a is odd and use induction with respect to m to prove that P has at least one complex root. For the general induction step, let α1,...,αn denote the roots of P over a splitting field of P.Fork ∈ Z, let Qk be the polynomial with roots αi + αj + kαiαj,1≤ i

3 ( / π) dθ θ The winding number can also be deﬁned by the integral 1 2 wr , where is the polar angle on R2 −{0}. (Observe that θ is multiple-valued, but dθ gives a well-deﬁned 1-form on R2 −{0}.) Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Web Site 95

induction hypothesis, and conclude that αi + αj + kαiαj is a complex number for some 1 ≤ i

Web Site

1. www.cs.amherst.edu/∼djv/fta.html Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Symmetries of . . . . . Regular Polygons . 9 . SECTION .

♣ We now go back again to ancient Greek mathematics, in which regular polygons played a central role. We learned that a regular n-sided polygon, Pn, n ≥ 3, can be represented by its vertices z(2kπ/n), k 0, 1,...,n− 1, on the complex plane C (Figure 9.1). By the multiplicative property of z(θ), θ ∈ R (see Section 5), we have

z(2kπ/n)z(2lπ/n) z(2(k + l )π/n), k, l ∈ Z.

We see that multiplication by z(2kπ/n) causes the index of the vertex to shift by k. More geometrically, the multiplicative property also tells us that multiplying complex numbers by z(θ) corresponds to the geometric transformation of counterclockwise rotation Rθ by angle θ around the origin (see Figure 9.2). In particular, the group

Figure 9.1 96 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

9. Symmetries of Regular Polygons 97

z(θ)z

z(θ)

Figure 9.2 of rotations

{R2kπ/n | k 0,...,n− 1}

(isomorphic with Zn) leaves the regular n-sided polygon invariant. Regularity thus implies that each vertex (and each edge) can be carried into any vertex (and any edge) by a suitable symmetry that leaves the polygon invariant. This seemingly innocent remark gives a profound clue to deﬁning regularity of polyhedra in space. We now ask the following more general question: What is the largest group of geometric transformations that leaves the regular n-sided polygon invariant? To answer this question, we need to make the term “geometric transformation” precise. Transfor- mation usually refers to a bijection of the ambient space, which, in this case, is the Cartesian plane R2. (A ﬁner point is whether we should require continuity or differentiability; fortunately, this is not essential here.) Since in Euclidean plane geometry we always work in the concrete model R2 (and avoid the headache of axiomatic treatment), we have the Euclidean distance function d : R2 × R2 → R, d(p, q) |p − q|,p,q∈ R2, and the concepts of angle, area, etc. The term “geometric” stands for transformations that preserve some of these geometric Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

98 9. Symmetries of Regular Polygons

quantities. In our present situation, we require the geometric transformation to preserve d; that is, we consider isometries. Explicitly, a transformation S : R2 → R2 is an isometry if d(S(p), S(q)) d(p, q), p, q ∈ R2. Euclid’s Elements tells us that an isometry also preserves angles, areas, etc. The set of all plane isometries form a group denoted by Iso (R2). (Actually, R2 equipped with d is usually denoted by E2. Since we have no fear of confusing R2 with other (non-Euclidean) models built on R2, we ignore this ﬁner point.) We now look at examples of planar isometries.

2 1. Rθ(p): rotation with center p ∈ R and angle θ ∈ R; 2 2. Tv: translation with vector v ∈ R ; 2 3. Rl: rection in a line l ⊂ R ; 1 4. Gl,v: glide rection along a line l with vector v (parallel to l); in fact, Gl,v Tv ◦ Rl Rl ◦ Tv. We now claim that every plane isometry is one of these. Al- though this result is contained in many textbooks, the proof is easy (especially in our model R2), so we will elaborate on it a little. Before the proof we assemble a few elementary facts. Given two lines l and l , the composition Rl ◦ Rl is a rotation if l and l intersect and a translation if l and l are parallel. The rotation angle is twice the signed angle from l to l; the translation vector is perpendicular to these lines, and its length is twice the signed distance from l to l. (Thus, rotations and translations are not all that different; consider a rotation and move the center “slowly to inﬁnity”; when the center leaves the plane, the rotation becomes a translation!) Notice that when decomposing a rotation with center at p as a product of rections Rl and Rl , the line l (or l) through p can be chosen arbitrarily. (What is the analogue of this for translations?) Let R2α(p) and R2β(q) be rotations with p q and let l denote the line through p and q. By the above, R2α(p) Rl ◦ Rl, where l is the unique line through p such that the angle from l to l is α.

1You may say, “I do not understand this!” and push this book away closing it. Well, you just performed a glide! Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

9. Symmetries of Regular Polygons 99

Similarly, R2β(q) Rl ◦ Rl , where l is the line through q such that the angle from l to l is β. The composition

R2α(p) ◦ R2β(q) (Rl ◦ Rl) ◦ (Rl ◦ Rl ) Rl ◦ Rl is the product of rections in the lines l and l.Ifα + β ∈ πZ, then l and l are parallel, and the composition is a translation with translation vector twice the vector from q to p.Ifα + β/∈ πZ, then the lines l and l intersect at a point r, and composition is a rotation with center r. It is convenient to write this rotation as R−2γ (r), since then

R2α(p) ◦ R2β(q) ◦ R2γ (r) I with α + β + γ ∈ πZ. This is due to W.F. Donkin (1851). If in the latter argument the roles of R2α(p) and R2β(q) are interchanged, −1 −1 then we obtain a rotation R2γ (s) R2α(p) ◦ R2β(q) with s r. In particular, the commutator −1 −1 R2α(p) ◦ R2β(q) ◦ R2α(p) ◦ R2β(q) is a translation. ♥ As a byproduct, we obtain that if a subgroup G ⊂ Iso (R2) contains no translations, then all the rotations in G have the same center. ♣ Finally, note that an isometry that fixes three noncollinear points is the identity, a fact that is easy to show. We are now ready to prove the claim. Let S : R2 → R2 be an isometry. Assume first that S is direct, that is, orientation preserving. We split the argument into two cases according to whether or not S has a fixed point. If S has a fixed point p ∈ R2, that is, S(p) p, then choose q ∈ R2 different from p. Let θ be the angle qpS(q). The −1 composition Rθ(p) ◦S leaves p and q fixed. It thus fixes every point −1 on the line through p and q. On the other hand, Rθ(p) ◦ S is direct (since S is), so that it must be the identity. We obtain S Rθ(p). Assume now that S has no fixed points. Let p ∈ R2 be arbitrary and consider the vector v emanating from p and terminating in S(p). −1 The composition (Tv) ◦ S is direct and leaves p fixed. By the first case, it is a rotation Rθ(p), i.e., S Tv ◦ Rθ(p). We claim that Rθ(p) is the identity, i.e., S Tv. ¬ Assume not. Arrange v to be the base of the isosceles triangle with vertex p opposite to v and angle θ at p as Figure 9.3 shows. Clearly, q is a fixed point of S Tv ◦ Rθ(p). ¬ Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

100 9. Symmetries of Regular Polygons

Figure 9.3 p

Second, assume that S is opposite, that is, orientation reversing.If S has a fixed point p, then let l be a line through p and consider the composition Rl ◦ S. This is a direct isometry that fixes p.Bythe previous case, it must be a rotation Rθ(p). We obtain Rl ◦ S Rθ(p), or equivalently, S Rl ◦ Rθ(p). We now write Rθ(p) Rl ◦ Rl , where l meets l at p and the angle between l and l is θ/2. We obtain S Rl ◦ Rl ◦ Rl Rl , so that S is a rection in a line. If S has no fixed point, then let p ∈ R2 be arbitrary and denote by q ∈ R2 the midpoint of the segment connecting p and S(p).Ifp, q and S(q) are collinear, it is easy to see that the line l through these points is invariant under S. Thus, the direct isometry Rl ◦ S keeps l invariant so that it is a translation Tv with translation vector v parallel to l. Thus Rl ◦ S Tv and so S Rl ◦ Tv Tv ◦ Rl Gl,v is a glide. If p, q and S(q) are not collinear then let r and s denote the orthogonal projections of p and S(p) to the line l through q and S(q) (Figure 9.4). We claim that s S(r). The triangle %qS(p)S(q) is isosceles, since d(p, q) d(q, S(p)) d(S(p), S(q)). Thus, the angles sqS(p) and sS(q)S(p) are equal. We obtain that the triangles %pqr and %S(p)S(q)s are congruent and oppositely oriented. But the same is true for %pqr and %S(p)S(q)S(r), so that s S(r) follows. Let Gl,v be the glide that sends %pqr to %S(p)S(q)S(r), where v has initial −1 point q and terminal point S(q). Then (Gl,v) ◦ S fixes p, q, and r, and so it is the identity. S Gl,v follows. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

9. Symmetries of Regular Polygons 101

rsSq (q)

S(p) Figure 9.4

Remark. Since rotations and translations are products of two rections, as a byproduct of the argument above we obtain that any planar isometry is the product of at most three rections.

We now go back to our regular polygons. Let X be a set (ﬁgure) in the plane and deﬁne the symmetry group of X as Symm (X) {S ∈ Iso (R2) | S(X) X}.

Theorem 3. For n ≥ 3, Symm (Pn) consists of the rotations

R2kπ/n R2kπ/n(0), k 0, 1,...,n− 1, n R ,...,R l ,...,l and reﬂections l1 ln in the lines 1 n joining the origin to the vertices and to the midpoints of the sides.

Proof. Let S ∈ Symm (Pn). It is clear that S can only be a rotation or a rection. Indeed, just look at Figure 9.5. Under S, vertices go to vertices and midpoints of sides go to midpoints of sides; in fact, S is a permutation on these two sets. Thus, the origin—the centroid of Pn—is left ﬁxed by S.(Forn even, the centroid is the midpoint of a diagonal connecting two vertices. For n odd, the centroid is on a line connecting the midpoint of a Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

102 9. Symmetries of Regular Polygons

Figure 9.5

side and the opposite vertex, and the center splits this segment in a speciﬁed ratio. What is this ratio?) If S is a rotation, then S Rθ Rθ(0) and θ 2kπ/n clearly follow. If S is rection in a line l, then l must go through the origin. Again it follows that l is one of the li’s, i 1,...,n.

Remark. As seen from the proof, there is a slight distinction between the structure of the lines l1,...,ln for n even and n odd. For example, take a look at the triangle P3 and the square P4 in Figure 9.6. We now take an algebraic look at the group Symm (Pn). Letting k a R2π/n, we see that a R2kπ/n, k 0,...,n− 1, so that the rotations generate the cyclic subgroup e a0,a,a2,...,an−1.

Figure 9.6 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

9. Symmetries of Regular Polygons 103

C b R b2 e b We denote this by n. Let l1 . Then since is a rection, and ba a−1b. (This needs veriﬁcation.) The rec- 2 n−1 tions b, ab, a b,...,a b are mutually distinct in Symm (Pn),so R ,...,R that they must give l1 ln (in a possibly permuted order). We obtain that Symm (Pn) is generated by two elements a and b and n 2 −1 relations a b e and ba a b. The elements of Symm (Pn) are e, a, a2,...,an−1; b, ab, a2b,...,an−1b.

This is called the dihedral group Dn of order 2n. The name comes from the fact that Dn is the symmetry group of a dihedron (as Klein called it), a spherical polyhedron with two hemispheres as faces and n vertices distributed uniformly along the common boundary. (Dn is also the symmetry group of the “reciprocal” spherical polyhedron with two antipodal vertices connected by n semicircles as edges that split the sphere into n congruent spherical wedges as faces.) Looking back, we see that studying the symmetries of regular polygons leads us to the cyclic group Cn of order n and the dihedral group Dn of order 2n. It is a remarkable fact that any ﬁnite subgroup of Iso (R2) is isomorphic to one of these.

Theorem 4. 2 2 Let G ⊂ Iso (R ) be a ﬁnite subgroup. Then G ﬁxes a point p0 ∈ R and is one of the following:

1. G is a cyclic group of order n, generated by the rotation R2π/n(p0). 2. G is a dihedral group of order 2n, generated by two elements: R2π/n(p0) and a reﬂection Rl in a line l through p0.

Proof. Let p ∈ R2 be any point and consider the orbit of G through p: G(p) {S(p) | S ∈ G}. This is a ﬁnite subset of R2 with elements listed as G(p) {p1,...,pm}. Each element in G is a permutation on this set. Now consider the centroid p +···+p p 1 m . 0 m Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

104 9. Symmetries of Regular Polygons

We claim that p0 is left ﬁxed by G. Let S ∈ G. We now use the classiﬁcation of plane isometries above, along with the fact that S permutes p1,...,pm, to conclude that S(p0) p0. To determine the structure of G, we split the proof into two cases.

(1) G contains only direct isometries. Since G ﬁxes p0, every element in G is a rotation Rθ(p0) with center p0. In what follows, we suppress p0. Let θ be the smallest positive angle of rotation. We claim that G is generated by Rθ. Since G is ﬁnite, it will then follow that G is cyclic. Indeed, let Rα ∈ G be arbitrary. The division algorithm tells us that

α mθ + β, m ∈ Z,

with remainder 0 ≤ β<θ. Since G is a group, Rβ Rα−mθ −m Rα ◦ (Rθ) ∈ G. Since θ is the smallest positive angle with Rθ ∈ G this is possible only if β 0. We obtain that α mθ, and hence

m Rα Rmθ (Rθ) .

n The rest is clear, since (Rθ) I for |G| n so that θ 2π/n.

(2) Assume that G contains an opposite isometry. Any opposite isometry with ﬁxed point p0 must be a rection Rl in a line l through p0. Thus, G contains rotations and rections, the former being a subgroup of G denoted by G+. By the previous case, G+ is generated by a rotation Rθ (with θ 2π/n). Let Rl ∈ G.As in the proof of Theorem 3, we have the following 2n elements in G:

2 n−1 I, Rθ,Rθ ,...,Rθ , 2 n−1 Rl,Rθ ◦ Rl,Rθ ◦ Rl,...,Rθ ◦ Rl.

These isometries are all distinct, and they form a subgroup G (of G) isomorphic with Dn. We must show that G G. It is enough to show that G contains all rections in G. Let Rl ∈ G, l l . Since k l and l intersect in p0, Rl ◦ Rl is a rotation in G, hence Rl ◦ Rl Rθ k for some k 0,...,n − 1. Thus Rl Rθ ◦ Rl, and this is listed above. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 105

Theorem 4 asserts in particular that the only possible groups of central symmetries in two dimensions are

C1,C2,C3,... and D1,D2,D3,.... Central symmetry frequently occurs in nature. Most of us observed in childhood that snowflakes have sixfold (some threefold) symmetries. Flowers usually have fivefold symmetries, and depending on whether the petals are bilaterally symmetric or not, their sym- mety group is D5 or only C5. We finish this section with a quotation from Hermann Weyl’s Symmetry regarding Theorem 4: “Leonardo da Vinci engaged in systematically determining the possible symmetries of a central building and how to attach chapels and niches without destroying the symmetry of the nucleus. In abstract modern terminology, his result is essentially our above table of the possible finite groups of rotations (proper and improper) in two dimensions.”

Problems

1. Show that the 3-dimensional cube ([0, 1]3 ⊂ R3) can be sliced by planes to obtain a square, an equilateral triangle, and a regular hexagon.

2. Prove that any two rotations Rθ(p) and Rθ(q) with the same angle θ ∈ R are conjugate in Iso (R2); that is, Rθ(q) Tv ◦ Rθ(p) ◦ T−v, where v is a vector from p to q.

3. (a) Let sn, n ≥ 3, denote the side length of Pn, the regular n-sided polygon inscribed in the unit circle. Show that 2 s2n 2 − 4 − sn.

Deduce from this that √ √ √ s4 2,s8 2 − 2,s16 2 − 2 + 2,....

Generalize these to show that √ s2n 2 − 2 + 2 +···+ 2,

with n − 1 nested square roots. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

106 9. Symmetries of Regular Polygons

(b) Let An be the area of Pn. Derive the formula √ n−1 n−1 A2n+1 2 s2n 2 2 − 2 + 2 +···+ 2,

with n − 1 nested square roots. (Hint: Half of the sides of Pn serve as heights of the 2n isosceles triangles that make up P2n, so that A2n nsn/2.) Conclude that √ lim 2n 2 − 2 + 2 +···+ 2 π, n→∞ where in the limit there are n nested square roots. In particular, we have √ lim 2 + 2 +···+ 2 2. n→∞ 4. (a) Prove that in the product of three rections, one can always arrange that one of the recting lines is perpendicular to both the others. (b) Derive Theorem 4 without the “orbit argument,” using the previously proved fact that all the rotations in G have the same center. 5. The reciprocal of a point p (a, b) ∈ R2 to the circle with center at the origin and radius r>0istheline given by the equation ax + by r2. Show that the reciprocals of the vertices z(2kπ/n), k 0,...,n− 1, of the regular n-sided polygon Pn to its inscribed circle give the sides of another regular n-sided polygon whose vertices are the midpoints of the sides of Pn.

Web Sites

1. www.maa.org 2. aleph0.clarku.edu/∼djoyce/java/elements/elements.html Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Discrete Subgroups . . . 2 . . of Iso (R ) . 10 . SECTION .

♣ In Section 9, we obtained a classification of all finite subgroups of the group of isometries Iso (R2) of R2 by studying symmetry groups of regular polygons. We saw that such subgroups cannot contain translations or glides, a fact that is intimately connected to boundedness of regular polygons. If we want to include translations and glides in our study, we have to start with unbounded plane figures and their symmetry groups. It turns out that classification of these subgroups is difficult unless we assume that the subgroup G ⊂ Iso (R2) does not contain rotations of arbitrarily small angle and translations of arbitrarily small vector length. (As we will see later, we do not have to impose any condition on rections and glides.) Groups G ⊂ Iso (R2) satisfying this condition are called discrete. In this section we give a complete classification of discrete subgroups of Iso (R2). Just as cyclic and dihedral groups can be viewed as orientation-preserving and full symmetry groups of regular polygons, we will visualize these groups as symmetries of frieze and wallpaper patterns. Thus, next time you look at a wallpaper pattern, you should be able to write down generators and relations for the corresponding symmetry group! Let G ⊂ Iso (R2) be a discrete group. Assume that G contains a translation Tv ∈ G that is not the identity (v 0). All powers 107 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

108 10. Discrete Subgroups of Iso (R2)

k of Tv are then contained in G (by the group property): Tv ∈ G, k k ∈ Z. Since Tv Tkv, we see that all these are mutually distinct. It follows that G must be infinite. (The same conclusion holds for glides, since the square of a glide is a translation.) We see that the presence of translations or glides makes G infinite. The following question arises naturally: If we are able to excise the translations from G, is the remaining “part” of G finite? The significance of an affirmative answer is clear, since we just classified all finite subgroups of Iso (R2). This gives us a good reason to look at translations first. Let T be the group of translations in R2. It is clearly a subgroup 2 of Iso (R ). From now on we agree that for a translation Tv ∈ T , we draw the translation vector v from the origin. Associating to Tv the vector v (just made unique) gives the map ϕ : T → R2, defined by 2 ϕ(Tv) v, v ∈ R . Since T ◦ T T ,v,v ∈ 2, v1 v2 v1+v2 1 2 R and −1 2 (Tv) T−v,v∈ R , we see that ϕ is an isomorphism. Summarizing, the translations in Iso (R2) form a subgroup T that is isomorphic with the additive group R2. Let G ⊂ Iso (R2) be a discrete group. The translations in G form a subgroup T G ∩ T of G. Since G is discrete, so is T. The isomorphism ϕ : T → R2 maps T to a subgroup denoted by 2 LG ⊂ R . This latter group is also discrete in the sense that it does not contain vectors of arbitrarily small length. By definition, LG is 2 the group of vectors v ∈ R such that the translation Tv is in G.We now classify the possible choices for LG.

Theorem 5. Let L be a discrete subgroup of R2. Then L is one of the following: Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

10. Discrete Subgroups of Iso (R2) 109

1. L {0}; 2. L consists of integer multiples of a nonzero vector v ∈ R2: L {kv | k ∈ Z}; 3. L consists of integral linear combinations of two linearly independent vectors v,w ∈ R2: L {kv + lw | k,l ∈ Z}.

Proof. We may assume that L contains a nonzero vector v ∈ R2. Let l R · v be the line through v. Since L is discrete, there is a vector in l ∩ L of shortest length. Changing the notation if necessary, we may assume that this vector is v. Let w be any vector in l ∩ L.We claim that w is an integral multiple of v. Indeed, w av for some a ∈ R since w is in l. Writing a k + r, where k is an integer and 0 ≤ r<1, we see that w−kv (a−k)v rv is in L. On the other hand, if r 0, then the length of rv is less than that of v, contradicting the minimality of v. Thus r 0, and w kv, an integer multiple of v. If there are no vectors in L outside of l, then we land in case 2 of the theorem. Finally, assume that there exists a vector w ∈ L not in l. The vectors v and w are linearly independent, so that they span a parallelogram P. Since P is bounded, it contains only ﬁnitely many elements of L. Among these, there is one whose distance to the line l is positive, but the smallest possible. By changing w (and P), we may assume that this vector is w. We claim now that there are no vectors of L in P except for its vertices. ¬ Assume the contrary and let z ∈ L be a vector in P. Due to the minimal choice of v and w, this is possible only if z terminates at a point on the opposite side of v or w. In the ﬁrst case, z − w ∈ L would be a vector shorter than v; in the second, z would be closer to l than w. ¬ Summariz- ing, we conclude that there are two linearly independent vectors v and w that span a parallelogram P such that P contains no vectors in L except for its vertices. Clearly, {kv + lw | k, l ∈ Z} is contained in L. To land in case 3 we now claim that every vector z in L is an Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

110 10. Discrete Subgroups of Iso (R2)

integral linear combination of v and w. By linear independence, z is certainly a linear combination z av + bw

of v and w with real coefﬁcients a, b ∈ R. We now write a k + r and b l + s, where k, l ∈ Z and 0 ≤ r, s < 1. The vector z − kv − lw rv + sw is in L and is contained in P. The only way this is possible is if r s 0 holds. Thus z kv + lw, and we are done.

We now return to our discrete group G ⊂ Iso (R2) and see that we have three choices for LG.IfLG {0}, then G does not contain any translations (or glides, since the square of a glide is a translation). In this case, G consists of rotations and rections only. By a result of the previous section, the rotations in G have the same center, say, p0. Since G is discrete, it follows that G contains only ﬁnitely many rotations. If Rl ∈ G is a rection, then l must go through p0, since otherwise, Rl(p0) would be the center of another rotation in G. Finally, since the composition of two rections in G is a rotation in G, there may be only at most as many rections in G as rotations (cf. the proof of Theorem 4). Summarizing, we obtain that if G is a discrete group of isometries with LG {0} then G must be ﬁnite. In the second case T, the group of all translations in G, is generated by Tv, and we begin to suspect that G is the symmetry group of a frieze pattern. Finally, in the third case T is generated by Tv and Tw, and T is best viewed by its ϕ-image 2 2 LG {kv + lw | k, l ∈ Z} in R . We say that LG is a lattice in R and G is a (2-dimensional) crystallographic group. Since any wallpaper pattern repeats itself in two different directions, we see that their symmetry groups are crystallographic. We now turn to the process of “excising” the translation part from G. To do this, we need some preparations. Recall that at the discussion of translations we agreed to draw the vectors v from 2 the origin so that the translation Tv by the vector v ∈ R acts on 2 p ∈ R by Tv(p) p + v. Now given any linear transformation 2 2 2 A : R → R (that is, A(v1 + v2) A(v1) + A(v2), v1,v2 ∈ R , and A(rv) rA(v), r ∈ R,v ∈ R2), we have the commutation rule Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

10. Discrete Subgroups of Iso (R2) 111

2 A ◦ Tv TA(v) ◦ A. Indeed, evaluating the two sides at p ∈ R ,we get

(A ◦ Tv)(p) A(Tv(p)) A(p + v) A(p) + A(v) and

(TA(v) ◦ A)(p) TA(v)(A(p)) A(p) + A(v).

Let O(R2) denote the group of isometries in Iso (R2) that leave the origin ﬁxed. O(R2) is called the orthogonal group. From the classiﬁcation of the plane isometries, it follows that the elements of O(R2) are linear.

Remark. ♠ We saw above that a direct isometry in O(R2) is a rotation 2 Rθ. These rotations form the special orthogonal group SO(R ),a 2 subgroup of O(R ). Associating to Rθ the complex number z(θ) establishes an isomorphism between SO(R2) and S1. Any opposite isometry in O(R2) can be written as a rotation followed by conjugation. Thus topologically O(R2) is the disjoint union of two circles.

♣ Occasionally, it is convenient to introduce superscripts ± to indicate whether the isometries are direct or opposite. Thus Iso +(R2) denotes the set of direct isometries in Iso (R2). Note that it is a subgroup, since the composition and inverse of direct isometries are direct. Iso−(R2) is not a subgroup but a topological copy of Iso+(R2). ♥ The elements of O(R2) are linear, so that the commutation rule above applies. We now define a homomorphism ψ : Iso (R2) → O(R2) as follows: Let S ∈ Iso (R2) and denote by v the vector that termi- −1 nates at S(0). The composition (Tv) ◦ S fixes the origin so that it 2 −1 is an element of O(R ). We define ψ(S) (Tv) ◦ S. To prove that −1 2 ψ is a homomorphism, we first write (Tv) ◦ S U ∈ O(R ),so that S Tv ◦ U. This decomposition is unique in the sense that if 2 2 S Tv ◦ U with v ∈ R and U ∈ O(R ), then v v and U U . −1 −1 Indeed, Tv ◦ U Tv ◦ U implies that (Tv ) ◦ Tv U ◦ U . The This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

112 10. Discrete Subgroups of Iso (R2)

right-hand side fixes the origin so that the left-hand side, which is a translation, must be the identity. Uniqueness follows. Using the notation we just introduced, we have ψ(S) U, where S T ◦ U S T ◦ U S T ◦ U v . Now let 1 v1 1 and 2 v2 2, where 2 2 v1,v2 ∈ R and U1,U2 ∈ O(R ). For the homomorphism property, we need to show that ψ(S2 ◦ S1) ψ(S2) ◦ ψ(S1). By definition, ψ(S1) U1 and ψ(S2) U2, so that the right-hand side is U2 ◦ U1. As for the left-hand side, we first look at the composition S ◦ S T ◦ U ◦ T ◦ U . 2 1 v2 2 v1 1 U U ◦ T Using the commutation rule for the linear 2, we have 2 v1 T ◦ U U2(v1) 2. Inserting this, we get S ◦ S T ◦ T ◦ U ◦ U . 2 1 v2 U2(v1) 2 1 Taking ψ of both sides amounts to deleting the translation part:

ψ(S2 ◦ S1) U2 ◦ U1. Thus ψ is a homomorphism. ψ is onto since it is identity on O(R2) ⊂ Iso (R2). The kernel of ψ consists of translations: ker ψ T . In particular, T ⊂ Iso (R2) is a normal subgroup. Having constructed ψ : Iso (R2) → O(R2), we return to our discrete group G ⊂ Iso (R2). The ψ-image of G is called the point-group of G, denoted by G¯ ψ(G). The kernel of ψ|G is all translations in G, that is, T. Thus, we have the following: ψ|G : G → G¯ ⊂ O(R2) and ker(ψ|G) T. ¯ For nontrivial LG, the point-group G interacts with LG in a beautiful way:

Theorem 6. ¯ G leaves LG invariant. This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

10. Discrete Subgroups of Iso (R2) 113

Proof. ¯ Let U ∈ G and v ∈ LG. We must show that U(v) ∈ LG. Since ¯ 2 U ∈ G, there exists S ∈ G, with S Tw ◦ U for some w ∈ R . The assumption v ∈ LG, means Tv ∈ G, and what we want to conclude, U(v) ∈ LG, means TU(v) ∈ G. We compute

−1 TU(v) TU(v) ◦ Tw ◦ (Tw)

−1 Tw ◦ TU(v) ◦ (Tw)

−1 −1 Tw ◦ U ◦ Tv ◦ U ◦ (Tw)

−1 S ◦ Tv ◦ S ∈ G, where the last but one equality is because of the commutation relation

U ◦ Tv TU(v) ◦ U as established above. The theorem follows.

G¯ is discrete in the sense that is does not contain rotations with arbitrarily small angle. This follows from Theorem 6 if LG is nontrivial. If LG is trivial, then by a result of the previous section, G is finite, and so is its (isomorphic) image G¯ under ψ. Since G¯ is discrete and fixes the origin, it must be finite! Indeed, by now this ¯ argument should be standard. Let Rθ ∈ G with θ being the smallest ¯ positive angle. Then any rotation in G is a multiple of Rθ. More- over, using the division algorithm, we have 2π nθ + r,0 ≤ r<θ, n ∈ Z, and r must reduce to zero because of minimality of θ. Thus θ 2π/n, and the rotations form a cyclic group of order n. Finally, there cannot be infinitely many rections, since otherwise their axes could get arbitrarily close to each other, and composing any two could give rotations of arbitrarily small angle. We thus accomplished our aim. G¯ gives a finite subgroup in O(R2) consisting of rotations and rections only. In particular, if LG {0}—that is, if G contains no nontrivial translations—then the kernel of ψ|G is trivial and so ψ|G maps G isomorphically onto G¯ . In particular, G is finite. By Theorem 4 of Section 9, G is cyclic or dihedral. This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

114 10. Discrete Subgroups of Iso (R2)

We are now ready to classify the possible frieze patterns, of which there are seven. According to Theorem 6, a frieze group G keeps the line c through LG invariant, and the group of translations T in G is an inﬁnite cyclic subgroup generated by a shortest translation, say, τ, in the direction of c. The line c is called the “center” of the frieze group. In addition to T, the only nontrivial direct isometries are rotations with angle π, called “half-turns,”and their center must be on c. The only possible opposite isometries are rection to c, rections to lines perpendicular to c, and glides along c.Inthe classiﬁcation below we use the following notations: If G contains a half-turn, we denote its center by p ∈ c.IfG does not contain any half-turns, but contains rections to lines perpendicular to c, the axis of rection is denoted by l, and p is the intersection point of n l and c. Otherwise p is any point on c. Let pn τ (p), n ∈ Z, and m the midpoint of the segment connecting p0 and p1. Finally, n let mn τ (m), the midpoint of the segment connecting pn and pn+1 (Figure 10.1). We are now ready to start. First we classify the frieze groups that contain only direct isometries: 1. G T 'τ(, so that G contains1 no half-turns, rections or glide rections. 2. G 'τ, Hp(. Aside from translations, G contains the half-turns n 2k τ ◦ Hp.Forn 2k even, τ ◦ Hp has center at pk, and for 2k+1 n 2k + 1 odd, τ ◦ Hp has center at mk. It is not hard to see that these are all the frieze groups that contain only direct isometries. We now allow the presence of opposite isometries.

2 3. G 'τ, Rc(. Since Rc I and τ ◦ Rc Rc ◦ τ, aside from T, this n group consists of glides τ ◦ Rc mapping p to pn.

c pmp Figure 10.1 1

1If G ⊂ Iso (R2), then 'G( denotes the smallest subgroup in Iso (R2) that contains G. We say that G generates 'G( (cf. “Groups” in Appendix B). This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

10. Discrete Subgroups of Iso (R2) 115

2 −1 4. G 'τ, Rl(. Since Rl I and Rl ◦ τ τ ◦ Rl, aside from T, G n consists of rections τ ◦ Rl. The axes are perpendicular to c, and according to whether n 2k (even) or n 2k + 1 (odd), the intersections are pk or mk. 5. G 'τ, Hp,Rc(. We have Hp ◦ Rc Rl ◦ Rc ◦ Rc Rl ∈ G.In n addition to this and τ, G includes the glides τ ◦ Rc (sending p n to pn) and τ ◦ Rl discussed above. 6. G 'τ, Hp,Rl (. Then l must intersect c perpendicularly at the midpoint of p and mk, for some k ∈ Z. 2 7. G 'Gc,v( is generated by the glide Gc,v with Gc,v τ.

Figure 10.2 depicts the seven frieze patterns. (The pictures were produced with Kali (see Web Site 2), written by Nina Amenta of the Geometry Center at the University of Minnesota.) Which corresponds to which in the list above? ¯ The fact that the point-group G leaves LG invariant imposes a severe restriction on G if LG is a lattice, the case we turn to next.

Figure 10.2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

116 10. Discrete Subgroups of Iso (R2)

R θ (v)

Figure 10.3 v

Crystallographic Restriction. Assume that G is crystallographic. Let G¯ denote its point-group. Then ¯ ¯ every rotation in G has order 1,2,3,4, or 6, and G is Cn or Dn for some n 1,2,3,4, or 6.

Proof. ¯ As usual, let Rθ be the smallest positive angle rotation in G, and let v ¯ be the smallest length nonzero vector in LG. Since LG is G-invariant, Rθ(v) ∈ LG. Consider w Rθ(v) − v ∈ LG (Figure 10.3). Since v has minimal length, |v|≤|w|. Thus, θ ≥ 2π/6,

and so Rθ has order ≤ 6. The case θ 2π/5 is ruled out since 2 Rθ (v) + v is shorter than v (Figure 10.4).

R θ (v)

2 R θ (v)

π/5 Figure 10.4 v Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

10. Discrete Subgroups of Iso (R2) 117

The first statement follows. The second follows from the classification of finite subgroups of Iso (R2) in the previous section.

Remark. For an algebraic proof of the crystallographic restriction, consider ¯ the trace tr (Rθ) of Rθ ∈ G,0<θ≤ π. With respect to a basis in LG, the matrix of Rθ has integral entries (Theorem 6). Thus, tr (Rθ) is an integer. On the other hand, with respect to an orthonormal basis, the matrix of Rθ has diagonal entries both equal to cos(θ).In particular, tr (Rθ) 2 cos(θ). Thus, 2 cos(θ) is an integer, and this is possible only for n 2, 3, 4, or 6.

Example If ω z(2π/n) is a primitive nth root of unity, then Z[ω]isa lattice iff n 3, 4, or 6. Indeed, the rotation R2π/n leaves Z[ω] invariant, since it is multiplication by ω. By the crystallographic restiction, n 3, 4, or 6. How do the tesselations look for n 3 and n 6?

The absence of order-5 symmetries in a lattice must have puzzled some ancient ornament designers. We quote here from Hermann Weyl’s Symmetry: “The Arabs fumbled around much with the number 5, but they were of course never able honestly to insert a central symmetry of 5 in their ornamental designs of double inﬁ- nite rapport. They tried various deceptive compromises, however. One might say that they proved experimentally the impossibility of a pentagon in an ornament.” Armed with the crystallographic restriction, we now have the tedious task of considering all possible scenarios for the point-group ¯ G and its relation to LG. This was done in the nineteenth century by Fedorov and rediscovered by Polya and Niggli in 1924. A description of the seventeen crystallographic groups that arise are listed as follows:

Generators for the 17 Crystallographic Groups 1. Two translations. 2. Three half-turns. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

118 10. Discrete Subgroups of Iso (R2)

3. Two rections and a translation. 4. Two parallel glides. 5. A rection and a parallel glide. 6. Rections to the four sides of the rectangle. 7. A rection and two half-turns. 8. Two perpendicular glides. 9. Two perpendicular rections and a half-turn. 10. A half-turn and a quarter-turn. 11. Rections in the three sides of a (π/4,π/4,π/2) triangle. 12. A rection and a quarter-turn. 13. Two rotations through 2π/3. 14. A rection and a rotation through 2 π/3. 15. Rections in the three sides of an equilateral triangle. 16. A half-turn and a rotation through 2π/3. 17. Rections is in the three sides of a (π/6,π/3,π/2) triangle.

Remark. The following construction sheds some additional light on the geometry of crystallographic groups. Let G be crystallographic and assume that G contains rotations other than half-turns. Let R2α(p) ∈ G,0<α<π/2, be a rotation with integral π/α (cf. the proof of Theorem 4 of Section 9). Let R2β(q) ∈ G,0<β<π/2, be another rotation with integral π/β such that d(p, q) is minimal. (R2β(q) exists since G is crystallographic.) Let l denote the line through p and q. Write R2α(p) Rl ◦ Rl, where l meets l at p and the angle from l to l is α. Similarly, R2β(q) Rl ◦Rl , where l meets l at q and the angle from l to l is β. Since α + β<π, the lines l and l intersect at a point, say, r. In fact, r is the center of the rotation −1 R2γ (r) (R2α(p) ◦ R2β(q)) Rl ◦ Rl . Since α, β, and γ are the interior angles of the triangle %pqr, we have α + β + γ πnthe other hand, since G is discrete, π/γ is rational. It is easy to see that minimality of d(p, q) implies that π/γ is integral. We obtain that α β γ + + , π π π 1

where the terms on the left-hand side are reciprocals of integers. Since π/α, π/β ≥ 3 (and π/γ ≥ 2), the only possibilities are α Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

10. Discrete Subgroups of Iso (R2) 119

Figure 10.5

β γ π/3; α β π/4,γ π/2; and α π/6,β π/3,γ π/2. (Which corresponds to which in the list above?)

As noted above, these groups can be visualized by patterns covering the plane with symmetries prescribed by the acting crystallographic group. Figure 10.5 shows a sample of four patterns (produced with Kali). Symmetric patterns2 date back to ancient times. They appear in virtually all cultures; on Greek vases, Roman mosaics, in the thirteenth century Alhambra at Granada, Spain, and on many other Muslim buildings.

2For a comprehensive introduction see B. Grunbaum¨ and G.C. Shephard, Tilings and Patterns, Freeman, 1987. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

120 10. Discrete Subgroups of Iso (R2)

To get a better view of the repetition patterns, we introduce the concept of fundamental domain. First, given a discrete group G ⊂ Iso (R2),afundamental set for G is a subset F of R2 which contains exactly one point from each orbit G(p) {S(p) | S ∈ G},p∈ R2.

A fundamental domain F0 for G is a domain (that is, a connected open set) such that there is a fundamental set F between F0 and 3 ¯ ¯ its closure F0; that is, F0 ⊂ F ⊂ F0, and the 2-dimensional area of ¯ the boundary ∂F0 F0 − F0 is zero. The simplest example of a fundamental set (domain) is given by the translation group G T 'Tv,Tw(. In this case, a fundamental domain F0 is the open parallelogram spanned by v and w.A fundamental set F is obtained from F0 by adding the points tv and tw,0≤ t<1. By the deﬁning property of the fundamental set, the “translates” S(F), S ∈ G, tile4 or, more sophisticatedly, tessellate R2. (Numerous tessellations appear in Kepler’s Harmonice Mundi, which appeared in 1619.) If a pattern is inserted in F, translating it with G gives the wallpaper patterns that you see. You are now invited to look for fundamental sets in Figure 10.5!

Problems

1. Prove directly that any plane isometry that ﬁxes the origin is linear. 2. Identify the frieze group that corresponds to the pattern in Figure 10.6. 3. Let L ⊂ R2 be a lattice. Show that half-turn around the midpoint of any two points of L is a symmetry of L. 4. Identify the discrete group G generated by the three half-turns around the midpoints of the sides of a triangle.

Figure 10.6

3See “Topology” in Appendix C. 4We assume that the tiles can be turned over; i.e., they are decorated on both sides. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Web Sites 121 Web Sites

1. www.geom.umn.edu/docs/doyle/mpls/handouts/node30.html 2. www.geom.umn.edu/apps/kali/start.html 3. www.math.toronto.edu/∼coxeter/art-math.html 4. www.texas.net/escher/gallery 5. www.suu.edu/WebPages/MuseumGaller/Art101/aj-webpg.htm 6. www.geom.umn.edu/apps/quasitiler/start.html Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Mobius¨ Geometry . . . 11 . SECTION .

♣ Recall from Section 7 that the stereographic projections hN : 2 2 2 2 S −{N}→R and hS : S −{S}→R combine to give −1 2 2 hN ◦ hS : R −{0}→R −{0}, where −1 2 2 N ◦ hS )(z) z/|z| 1/z,¯ 0 z ∈ C R . Strictly speaking, this is not a transformation of the plane since it is undeﬁned at z 0. Under hN , however, it corresponds to the transformation of the unit sphere S2: −1 −1 −1 2 2 hN ◦ N ◦ hS ) ◦ hN hS ◦ hN : S → S which sends a spherical point p (a,b,c)to (a, b, −c) so that it is spherical rection in the equatorial circle S1 of S2! That this holds can be seen from Figure 11.1. Rections in lines play a central role in Euclidean geometry. In fact, we saw in Section 9 that every isometry of R2 is the product of at most three rections. In spherical geometry the ambient space is S2, lines are great circles of S2, and rections are given by spatial rections in planes in R3 spanned by great circles. For example, −1 2 2 2 hS ◦ hN : S → S is the restriction to S of the spatial rection 122 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

11. Mobius¨ Geometry 123

(a, b, c)

(a, b, −c)

S Figure 11.1

(a,b,c) → (a, b, −c) in the coordinate plane R2 ⊂ R3 spanned by the first two axes. We could now go on and study isometries of S2 and develop spherical geometry in much the same way as we developed plane geometry. Instead, we put a twist on this and insist that we want to view all spherical objects in the plane! “Viewing”, of course, means not only the visual perception, but also the description of these objects in terms of Euclidean plane concepts. This is possible by the 2 2 stereographic projection hN : S −{N}→R , which does not quite map the entire sphere to R2 (this is impossible), but leaves out the North Pole N. Our spherical rections will thus become “singular” when viewed on R2. Tocircumvent this difficulty, we attach an “infinite point” ∞ to R2 and say that the singular point of the transformation must correspond to ∞. For example, the transformation z → 1/z¯ is singular at the origin but becomes well defined on the extended plane Rˆ 2 R2 ∪{∞}, where we agree that it should send 0to∞. Of course Rˆ 2 is nothing but S2, but it is much easier to view the action on R2 plus one point than on S2. Our path is now clear; we need to consider all finite compositions of spherical rections 2 2 in great circles of S and pull them down to R (via hN ). The transformations of R2 (or rather Rˆ 2) obtained this way are named after Mobius.¨ (The concept of rection in a circle in Rˆ 2, the case of a single spherical rection, was invented by Steiner around 1828.) Although this project is not difficult to carry out, we will pursue a different track and work in Rˆ 2 from the beginning. Our starting Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

124 11. Mobius¨ Geometry

Figure 11.2

point is the following observation: Consider the equatorial circle 1 2 3 1 S ⊂ R ⊂ R . Under hN , it corresponds to itself. Now rotate S around the ﬁrst axis in R3 by various angles (Figure 11.2). Under hN , the rotated great circles correspond to various circles and, when the great circle passes through N, to a straight line! The conclusion is inevitable. In Mobius¨ geometry we have to treat circles and lines on the same footing. Thus when we talk about circles we really mean circles or lines. For the farsighted this is no problem; a line on the extended plane Rˆ 2 becomes a circle by closing it up with ∞! Analytically, however, circles and lines have different descriptions:

2 Sr(p0) {p ∈ R | d(p, p0) r}, and

2 lt(p0) {p ∈ R | p · p0 t}∪{∞},

2 where Sr(p0) ⊂ R is the usual Euclidean circle with center p0 and radius r, and lt(p0) is the usual Euclidean line (extended with ∞) with normal vector p0 (· is the dot product). Notice that by ﬁxing p0 and varying r and t, we obtain concentric circles and parallel lines (Figure 11.3).

Figure 11.3 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

11. Mobius¨ Geometry 125

ˆ 2 ˆ 2 To define the rection RS : R → R in a circle S Sr(p0) in general, we rely on the special case where RS is obtained from 2 2 2 a spherical rection RC : S → S in a great circle C ⊂ S , i.e., −1 where RS hN ◦ RC ◦ hN . Since hN (C) S, this is precisely the case where S ∩ S1 contains an antipodal pair of points. At the end of Section 7, we proved that hN is circle-preserving. Since this is automatically true for isometries (such as RC), we see that RS is also circle-preserving in our special case. We now define the rection RS in a general circle S Sr(p0) by requiring that (i) RS should be circle-preserving, (ii) it should fix each point of S, and (iii) RS should interchange p0 and ∞. That these conditions determine RS uniquely follows by taking a careful look at the construction in Fig- ure 11.4, in which the point p is chosen outside of S. (The segment connecting p and q is tangent to S at q.) When p is inside S, the points p and RS(p) should be interchanged. With this in mind, our argument is the following. The line l through p and p0 should be mapped by RS to itself, since it contains both p0 and ∞ (which are interchanged), and it also goes through two diametrically opposite points of S (which stay fixed). Thus if p is on l then so is RS(p). The circle S through q, q and p0, and the line l through q, q (and ∞)are interchanged by RS since they both contain q, q (that stay fixed). Thus, if p is outside of S, then RS(p) must be the common intersection of the lines l and l. As noted above, the situation is analogous when p is inside S with the roles of p and RS(p) interchanged.

l r p 0 RS (p) l p

q Figure 11.4 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

126 11. Mobius¨ Geometry

The analytical formula for RS can also be read off from Figure 11.4. We have r 2 ˆ 2 RS(p) p0 + (p − p0), p ∈ R , d(p, p0)

with p0 and ∞ corresponding to each other. To see that this is true, we ﬁrst note that p0, p, and RS(p) are collinear, so that RS(p) − p0 λ(p − p0) for some λ ∈ R to be determined. We have d(R (p), p ) λ S 0 . d(p, p0) The angles qpp0 and p0qRS(p) are equal (Euclid!), so that the triangles %qpp0 and %p0qRS(p) are similar. Thus, d(R (p), p ) r S 0 , r d(p, p0) so that d(R (p), p ) r 2 λ S 0 , d(p, p0) d(p, p0)

and the formula for RS follows when p is outside S. Instead of checking our computations for the case where p is inside S, it sufﬁces to 2 2 show that RS is the identity. To work out RS, we ﬁrst note that r 2 RS(p) − p0 (p − p0), d(p, p0) so that r 2 r2 d(RS(p), p0) d(p, p0) . d(p, p0) d(p, p0) We now compute r 2 2 RS(p) p0 + (RS(p) − p0) d(RS(p), p0) r 2 r 2 p + (p − p ) 0 2 0 r /d(p, p0) d(p, p0)

p0 + p − p0 p. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

11. Mobius¨ Geometry 127

R (p) l

l t (p0 ) p Figure 11.5

2 Thus, RS is the identity, as it should be for a true rection. The story is the same for the rection Rl in the line lt(p0) with normal vector p0. The explicit formula is

2(p · p0 − t)p0 Rl(p) p − , 2 |p0| and Rl(∞) ∞(Figure 11.5). To show this we note ﬁrst that Rl(p) − p must be parallel to the normal vector p0 of lt(p0); that is,

Rl(p) − p λp0.

We determine λ using the fact that lt(p0) is the perpendicular bisector of the segment connecting p and Rl(p). Analytically, the midpoint

p + Rl(p) 2 of this segment must be on lt(p0):

(p + Rl(p)) · p0 2t.

On the other hand, Rl(p) p + λp0, so that (t − p · p ) λ 2 0 2 |p0| and the formula for Rl follows. We now define the Mobius¨ group Möb (Rˆ 2) as the group of all finite compositions of rections in circles and lines. Aside from Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

128 11. Mobius¨ Geometry

this being a group, one more fact is clear. Our plane isometries are contained in the Mobius¨ group Iso (R2) ⊂ Möb (Rˆ 2). This is because every plane isometry is the composition of (at most three) rections in lines. (Note also that we automatically extended the plane isometries to Rˆ 2 by declaring that they should send ∞ to itself.) Before we go any further, note that any spherical isometry of S2 is the composition of spherical rections. (The proof is analogous to the planar case.) Thus, conjugating the group Iso (S2) of all spherical isometries by the stereographic projection hN , we obtain 2 −1 ˆ that hN ◦ Iso (S ) ◦ hN is a subgroup of Möb (C)! Finally, notice that 2 2 −1 Iso (R ) and hN ◦ Iso (S ) ◦ hN intersect in the orthogonal group O(R2) (cf. Section 10). ◦ ♣ Let k>0. Let RS be rection in S1(0) and RS rection in √ ◦ S k(0). The composition RS ◦ RS works out as follows: ◦ p (RS ◦ R )(p) RS S |p|2 √ k 2 p kp. d(0,p/|p|2) |p|2 ◦ We see that RS ◦ RS is nothing but central dilatation with ratio of magnification k>0. We introduced rections in circles by the requirement that they should be circle-preseving. This enabled us to derive an explicit formula for RS as above. It is, however, not clear whether RS is actually circle-preserving. Our next result states just this.

Theorem 7. Let S be any M¨obius transformation on Rˆ 2. Then S maps circles to circles.

Proof. Since isometries and central dilatations send circles to circles, we S S (p ) (T )−1 ◦ may assume that is rection in a circle r 0 . Since p0 S ◦ T S ( ) p p0 is rection in the circle r 0 , we may assume that 0 0. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

11. Mobius¨ Geometry 129

T Replacing p0 in the previous argument with central dilatation with ◦ ratio r>0, we may assume that S RS, rection in the unit circle S1(0). Now let I be any circle. I can be described by the equation 2 2 α|p| − 2p · p0 + β 0,p∈ R , 2 where α, β ∈ R and p0 ∈ R . The choice α 1 gives the circle with center at p0, and α 0 deﬁnes the line with normal vector p0. 2 ◦ 2 Dividing by |p| and rewriting this in terms of q RS(p) p/|p| , we obtain 2 α − 2q · p0 + β|q| 0, and this is the equation of another circle. We are done.1

Remark. ♥ Mobius¨ transformations generalize to any dimensions. In fact, n the deﬁning formula for the rection RS works for R if we think n of d as the Euclidean distance function on R .Here,S Sr(p0) is the (n − 1)-dimensional sphere with center p0 and radius r.Asan interesting connection between Mobius¨ geometry and the stereographic projection hN , we note here that, for n 3, hN is nothing R S2 ⊂ 3 S S√ (N) but the restriction of S to R , where √ 2 is the sphere with center at the North Pole N and radius 2. This can be worked out explicitly by looking at Figure 11.6. (Do you see the Lune of 2 Hippocrates here?) In fact, all we need to show is that RS(p), p ∈ S , 2 is in R , or equivalently that RS(p) and N are orthogonal. Taking dot products and using |p| 1, we compute 2 RS(p) · N 1 + (p − N) · N |p − N|2 2 1 + (p · N − 1) 0. 2 − 2p · N

1The first part of the proof was to reduce the case of general Mobius¨ transformations to the single case of rection in the unit circle. Systematic reduction is a very useful tool in mathematics. Here’s a joke: A mathematician and a physicist are asked to solve two problems. The first problem is to uncork a bottle of wine and drink the contents. They solve the problem the same way: They both uncork the bottles and drink the wine. The second problem is the same as the first, but this time the bottles are open. The physicist (without much hesitation) drinks the second bottle of wine. The mathematician puts the cork back in the bottle and says, “Now apply the solution to the first problem!” Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

130 11. Mobius¨ Geometry

R (p) S

Figure 11.6 We can be even bolder and realize that no computation is needed if we accept the generalization of Theorem 7 to dimension 3, since 2 RS maps S to a “sphere” that contains ∞ RS(N), so that (looking S2 ∩ S√ (N) S2 R at the ﬁxed point set 2 ) the image of under S must be R2.

Problems

1. Show that the Mobius¨ group is generated by isometries, dilatations with center at the origin, and one rection in a circle. 2. Generalize the observation in the text and show that the composition of re- ﬂections in two concentric circles with radii r1 and r2 is a central dilatation 2 with ratio of magniﬁcation (r1/r2) . Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

. . . . . Complex Linear . . . . . Fractional . . . 12 . SECTION . Transformations

♣ In Section 11 we created the Mobius¨ group by playing around with stereographic projections, which were a fundamental ingre- dient in our proof of the FTA. Taking a closer look at the proof, however, we see that we have not attained full understanding of M¨ob (Rˆ 2). For example, in the proof of the FTA, it is a crucial fact that rection to the unit circle can be written not only as z → z/|z|2, but also as z → 1/z¯ (only the latter gave smoothness of the polynomial P on S2 across the North Pole). The key to better understanding of the Mobius¨ group is to introduce what is called “complex language in geometry.”As usual, we start modestly and express the basic geometric transformations in terms of complex variables:

2 1. Translation Tv by v ∈ C (considered as a vector in R ) can be deﬁned as Tv(z) z + v. 2. Rotation Rθ with center at the origin and angle θ ∈ R is nothing but multiplication by z(θ) cos θ + i sin θ; that is, Rθ(z) z(θ)z, z ∈ C. Now the rotation Rθ(p0) with center p0 ∈ C can be written R (p ) T ◦R ◦(T )−1 R (p )(z) R (z−p )+p as θ 0 p0 θ p0 , so that θ 0 θ 0 0 z(θ)(z − p0) + p0. 3. Rection Rl in a line can be written in terms of complex variables as follows: Let the line l lt(p0) be given by !(zp¯ 0) t, 131 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

132 12. Complex Linear Fractional Transformations

p0 0. (The dot product of two complex numbers z and w considered as plane vectors is given by the real part !(zw)¯ (zw¯ +¯zw)/2. This is what we just used here.) Now, by Section 11, Rl can be written as

2(!(zp¯ 0) − t)p0 zp¯ 0 +¯zp0 − 2t Rl(z) z − z − 2 |p0| p¯ 0 p z¯ − t − 0 2 . p¯ 0

4. The complex expression for the glide Gl,v Tv ◦ Rl Rl ◦ Tv follows from the expressions of Tv and Rl above. 5. Rection RS to the circle Sr(p0), r>0, is given by

2 2 2 r p0z¯ + (r −|p0| ) RS(z) + p0 . z¯ − p¯ 0 z¯ − p¯ 0 Comparing (3) and (5), we see something in common! They are both of the form az¯ + b z → ,ad− bc . cz¯ + d 0

(In the ﬁrst case the “determinant” ad − bc works out to be −p0p¯ 0 2 2 2 2 −|p0| 0, and in the second, −p0p¯ 0 − (r −|p0| ) −r 0.)

Theorem 8. Let Möb+(Cˆ ) denote the group of direct (orientation-preserving) Möbius transformations of the extended plane Cˆ . Then Möb+(Cˆ ) is identical to the group of complex linear fractional transformations az + b z → ,ad− bc , a,b,c,d,∈ . cz + d 0 C

Proof. Möb+(Cˆ ) is the subgroup of Möb (Cˆ ) consisting of those Mobius¨ transformations that are compositions of an even number of re- flections to lines and circles. As noted above, a rection has the form az¯ + b z → ,ad− bc , a,b,c,d ∈ . cz¯ + d 0 C Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

12. Complex Linear Fractional Transformations 133

The composition of two such rections (with different a’s, b’s, c’s, and d’s) is clearly a linear fractional transformation (cf. also Problem 1), so we have proved that M¨ob+(Cˆ ) is contained in the group of linear fractional transformations. For the converse, let g be a linear fractional transformation az + b g(z) ,ad− bc . cz + d 0

If c 0, then g(z) (a/d)z + (b/d), so that if a d, then g is a translation, and if a d, then g is the composition of a translation, a rotation, and a central dilatation. This latter statement follows from rewriting g as a g(z) (z − p) + p, d where p (b/d)/(1−(a/d)), and then looking at the complex form of rotations. (a/d |a/d|·z(θ), where θ arg(a/d) and |a/d| gives the ratio of magnification.) Thus, we may assume c 0. To begin with this case, we first note that z → 1/z is in Möb (Cˆ ) since it is rection to the unit circle followed by conjugation (rection to the real axis). We now rewrite g as bc − ad a g(z) + c2(z + d/c) c and conclude that g is the composition of a translation, the Mobius¨ transformation z → 1/z, a rotation, a central dilatation, and finally another translation. Thus g ∈ Möb (Cˆ ), and we are done.

Remark. More geometric insight can be gained by introducing the isometric circle Sg of a linear fractional transformation az + b g(z) ,ad− bc , a,b,c,d ∈ , cz + d 0 C as

1/2 Sg {z ∈ C ||cz + d| |ad − bc| }. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

134 12. Complex Linear Fractional Transformations

(Here we assume that c 0.) That Sg is a circle is clear. The name “isometric” comes from the fact that if z, w ∈ Sg, then we have

az + b aw + b |g(z) − g(w)| − cz + d cw + d

(ad − bc)(z − w) |z − w|, (cz + d)(cw + d)

so that g behaves on Sg as if it were an isometry! Let R√S denote the rection in Sg. Since Sg has center −d/c and radius |K|/|c|, where K ad − bc, we have |K|/|c|2 d RS(z) − z¯ + d/¯ c¯ c |K| d − . c(cz + d) c

The composition g ◦ RS is computed as

aRS(z) + b (g ◦ RS)(z) cRS(z) + d a(cR (z) + d) − (ad − bc) S c(cRS(z) + d) a|K|/(cz + d) − K

c|K|/(cz + d) K a − (cz + d) + . c|K| c

This map is of the form z → pz¯ + q with |p| 1. It can be decomposed as z →¯z → pz¯ → pz¯ + q.

The first arrow represents rection to the first axis; the second, rotation by angle arg p around the origin; the third, translation by q. The second and third are themselves compositions of two rections. So we find that z → pz¯ + q, |p| 1, is the composition of an odd number of at most five rections. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

12. Complex Linear Fractional Transformations 135

Summarizing, we ﬁnd that g ◦ RS is the composition of an odd number of rections, so that g is the composition of an even number of rections.

As a corollary to Theorem 7, we ﬁnd that linear fractional transformations are circle-preserving. We will return to this later. Given a direct Mobius¨ transformation represented by the linear fractional transformation az + b z → ,ad− bc , cz + d 0 multiplying a, b, c, d by the same complex constant represents the same Mobius¨ transformation. Choosing this complex constant suitably, we can attain ad − bc 1. We now introduce the complex special linear group SL(2, C), consisting of complex 2 × 2-matrices with determinant 1. ♥ By the above, we have + M¨ob (Cˆ ) ∼ SL(2, C)/{±I}. This complicated-looking isomorphism (see Problems 1 and 2) means that each Mobius¨ transformation can be represented by a matrix A in SL(2, C), with A and −A representing the same Mobius¨ transformation. If we write A as

ab A , cd then the Mobius¨ transformation is given by the linear fractional transformation az + b z → ,z∈ ˆ . cz + d C In what follows, we will not worry about the ambiguity caused by the choice in ±A; our group to study is SL(2, C). ♣ To illustrate how useful it is to represent direct Mobius¨ transformations by linear fractional transformations, we now show that ˆ ˆ given two triplets of distinct points z1, z2, z3 ∈ C and w1,w2,w3 ∈ C, there is a unique direct Mobius¨ transformation that carries z1 to w1, Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

136 12. Complex Linear Fractional Transformations

z2 to w2, and z3 to w3. Clearly, it is enough to show this for w1 1, w2 0, and w3 ∞. If none of the points z1, z2, z3 is ∞, then the Mobius¨ transformation is given by ! z − z z − z z → 2 1 2 z − z3 z1 − z3

If z1, z2,orz3 ∞, then the transformation is given by z − z z − z z − z 2 , 1 3 , 2 . z − z3 z − z3 z1 − z2 As for unicity, assume that a direct Mobius¨ transformation az + b z → ,ad− bc , cz + d 1 fixes 1, 0, and ∞. Since 0 is fixed, b 0. Since ∞ is fixed, c 0. Now 1 is fixed, so that 1 (a/d) and we end up with the identity. We finally note that any opposite Mobius¨ transformation can be written as az¯ + b z → ,ad− bc , a,b,c,d ∈ . cz¯ + d 1 C Indeed, if g is an opposite Mobius¨ transformation, then z → g(z)¯ gives an element of Möb+(Cˆ ), a linear fractional transformation. One final note: Mobius¨ transformations are not only circle- preserving but angle-preserving as well. What do we mean by this? An angle, after all, consists of two half-lines meeting at a point. An arbitrary transformation maps an angle to two curves that join at the image of the meeting point. The answer, as you know from calculus, is that two curves that meet at a point also have an angle defined by the angle of the corresponding tangent vectors at the meeting point. We now turn to the proof of preservation of angles (termed conformality later) under Mobius¨ transformations. This follows from conformality of the stereographic projection. We give here an independent analytic proof. (For a geometric proof, see Problem 8.) Since isometries preserve angles (congruent triangles in Euclid’s Elements), we can reduce the case of general Mobius¨ transformations to the case of rections in circles. Since central dilatations preserve angles (Euclid’s Elements again on similar triangles), the Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 137 only case we have to check is rection to the unit circle S1 ⊂ C. This is given by z → 1/z¯,0 z ∈ C. Even conjugation can be left out, since it is an isometry. Thus, all that is left to check is that z → 1/z preserves angles. ♦ Let 0 z0 ∈ C and γ : (−a, a) → C, a>0, a smooth curve with γ(0) z0. The Mobius¨ transformation z → 1/z sends the tangent vector γ(0) to the tangent vector (d/dt)(1/γ(t))t 0. The latter computes as d γ(t) γ ( ) 1 − − 0 dt γ(t) γ(t)2 z2 t 0 t 0 0 since γ(0) z0. Thus z → 1/z acts on the vector γ (0) by multiply- − /z2 ing it with the constant factor 1 0. This is central dilatation by /|z |2 (− /z2) π− z ratio 1 0 followed by rotation by angle arg 1 0 2arg 0. Both of these preserve angles, and we are done. If we accept that Mobius¨ transformations in R3 are angle- preserving (Problem 8 extended to R3), then the construction in the remark at the end of Section 11 shows that the stereographic projection is also angle-preserving. This we proved in Section 7 directly. We are now ready to introduce hyperbolic plane geometry in the next section.

Problems

1. Show that composition of linear fractional transformations corresponds to matrix multiplication in SL(2, C). 2. Check that the inverse of the linear fractional transformation z → (az + b)/(cz + d), ad − bc 0, is w → (dw − b)/(−cw + a). 3. Find a linear fractional transformation that carries 0 to 1, i to −1, and −i to 0. ˆ 4. Let z1,z2,z3,z4 ∈ C with z2,z3,z4 distinct. (a) Find a linear fractional transformation f that carries z2 to 1, z3 to 0, and z4 to ∞. ˆ (b) Deﬁne the cross-ratio (z1,z2,z3,z4) as f(z1) ∈ C. Show that the cross-ratio is invariant under any linear fractional transformation; that is, if g is a linear fractional transformation, we have (g(z1), g(z2), g(z3), g(z4)) (z1,z2,z3,z4). (c) Prove that (z1,z2,z3,z4) ∈ R iff z1,z2,z3,z4 lie on a circle or a straight line. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

138 12. Complex Linear Fractional Transformations

(d) Using (a)–(c), conclude that linear fractional transformations are circle- preserving. 5. Find a linear fractional transformation that maps the upper half-plane {z ∈ C |"(z) > 0} to the unit disk {z ∈ C ||z| < 1}. 6. Describe the Mobius¨ transformation z → 1 +¯z, z ∈ Cˆ , in geometric terms. 7. Show that the non-Mobius¨ transformation z → z + 1/z, z ∈ Cˆ , maps circles with center at the origin to ellipses. What is the image of a half-line emanating from the origin? 8. Show that Mobius¨ transformations preserve angles in the following geometric way: (a) Consider a single rection RS in a circle S Sr (p0). Use the identity 2 d(p, p0)d(RS(p), p0) r

(Figure 11.4) to show that RS maps circles orthogonal to S to themselves. (b) Given a ray emanating from a point p not in S, construct a circle that is orthogonal to the ray at p, and also orthogonal to S at the intersection points. (c) Apply (b) to the two sides of an angle, and notice that two intersecting circles meet at the same angle at their two points of intersection. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

. . . . . “Out of Nothing I . . . . . Have Created a New . . . 13 . SECTION . Universe”—Bolyai

♣ Plane isometries have one characteristic property that, although obvious, has not been mentioned so far. They map straight lines to straight lines! The more general Mobius¨ transformations do not have this property, but we have seen that they map circles (in the general sense; i.e., circles and lines) to circles. Thus the question naturally arises: Does there exist a “geometry” in which “lines” are circles and “isometries” are Mobius¨ transformations? The answer is not so easy if we think of the strict guidelines, called the Five Pos- tulates, that Euclid gave around 300 b.c.intheElements to obtain a decent geometry. Although we again shy away from the axiomatic treatment, we note here that for any geometry it is essential that through any two distinct points there be a unique line. The famous Fifth Postulate (actually, the equivalent Playfair’s axiom), Through a given point not on a given line there passes a unique line not intersecting the given line. is another matter. Even Euclid seemed reluctant to use this in his proofs. Returning to our circles and Mobius¨ transformations, we now have a general idea how to build a geometry in which “lines” are circles. We immediately encounter difﬁculty, since looking at the 139 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

140 13. “Out of Nothing I Have Created a New Universe”—Bolyai

Figure 13.1

pencil of circles in Figure 13.1 we see that the required unicity of the “line” (that is a circle) through two distinct points fails. There are too many ‘lines’ in our geometry! This is because a circle is determined by three of its distinct points, not two. Thus, to ensure unicity, we have to rule out a lot of circles. This can be done elegantly by imposing a condition on our circles. One simple idea is to fix a point on the plane and require all circles to pass through this point. This point cannot be in the model. Assuming that it is the origin, we perform the transformation z → 1/z, z ∈ Cˆ , and realize that our “circles” become straight lines! (This is because this transformation, like any Mobius¨ transformation, is circle-preserving, so that circles through the origin transform into “unbounded circles” that are straight lines.) We thus rediscover Euclidean plane geometry. A somewhat more sophisticated idea is to fix a line l and consider only those circles that intersect l perpendicularly. The points on this line cannot be in our model. l splits the plane into two half-planes of which we keep only one, denoted by H2, where the superscript indicates the dimension. That this works can be seen as follows: Take two distinct points p and q in H2. If the line through p and q is perpendicular to l, then we are done. Otherwise, the perpendicular bisector of p and q intersects l at a point c. Now draw a circle through p and q with center c. Since we keep only H2 as the point-set of our geometry, we see that the “line” through p and q is, in the first case, the vertical half-line through p and q (ending at a point in l not in H2) and, in the second case, the semicircle through p and q perpendicular to l at its endpoints. The location Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

13. “Out of Nothing I Have Created a New Universe”—Bolyai 141 of l is irrelevant, so we might just as well take it the real axis. We choose H2 to be the upper half-plane. We thus arrive at the half- plane model of hyperbolic geometry, due to Poincare.´ In this model hyperbolic lines are Euclidean half-lines or semi-circles perpendicular to the boundary of H2. The boundary points of H2 are called ideal or (for the obvious reason) points at infinity. It is easy to see that this geometry satisfies the first four postulates of Euclid, along with unicity of the hyperbolic line through two distinct points. What about the Fifth Postulate? In Figure 13.2 four hyperbolic lines meet at a point, and they are all parallel to the vertical hyperbolic line. The Fifth Postulate thus fails in this model. We have accomplished what mathematicians have been unable to do for almost 2000 years—independence of the Fifth Postulate from the first four! (In fact, we are concentrating here on the explicit model too much. One of the greatest discoveries of Bolyai was the invention of “absolute geometry” in the axiomatic treatment.) This geometry (although not this particular model) was discovered about 1830 by Bolyai (1802–1860) and Lobachevsky (1793–1856), and the story involving Gauss is one of the most controversial pieces in the history of mathematics.1 We now return to Mobius¨ transformations. Our path is clear; we need to see which Mobius¨ transformations map H2 onto itself. No- tice that a Mobius¨ transformation of this kind automatically maps the real axis to itself so that, being circle-preserving and conformal ( angle-preserving), it will automatically map our hyperbolic

Figure 13.2

1For a good summary, see M.J. Greenberg, Euclidean and non-Euclidean Geometries, Development and History, Freeman, 1993. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

142 13. “Out of Nothing I Have Created a New Universe”—Bolyai

lines to hyperbolic lines (since it preserves perpendicularity at the boundary points). We ﬁrst treat the case of direct Mobius¨ transformations, viewed as elements of SL(2, C).

Theorem 9. A matrix in SL(2,C) leaves H2 invariant iff its entries are real; that is, the matrix belongs to SL(2,R).

Proof. Let g be a linear fractional transformation az + b g(z) ,ad− bc , a,b,c,d ∈ , cz + d 1 C and assume that g maps H2 onto itself. Then g maps the real axis −1 onto itself. In particular, g (0) r0 is real or ∞. We assume that r0 ∈ R since r0 ∞can be treated analogously. The composition g ◦ T r0 ﬁxes the origin. It has the form a(z + r ) + b az + b + ar (g ◦ T )(z) 0 0 . r0 c(z + r0) + d cz + d + cr0

The numbers a, b + ar0, c, d + cr0 are real iff a, b, c, d are. Thus, it is g ◦ T enough to prove the theorem for r0 . Since this transformation fixes the origin, by changing the notation we can assume that the original linear fractional transformation g fixes the origin. This means that b 0 and az g(z) ,ad . cz + d 1 First assume that c 0, so that g(z) (a/d)z. Since g maps the real axis onto itself, a/d is real. Since ad 1, it follows that a/d a2 is real. This means that a is real or purely imaginary. In the first case we are done. In the second, a ia0 with a0 real. Thus, d /a −i/a g(z) (a/d)z −a2z i ∈ H2 1 0 and 0 . This maps outside of H2, so that this case is not realized. Next, we assume that c 0. Taking z r real and working out ar/(cr + d), it follows easily that this latter fraction is real iff a/c and a/d are both real. As before a/d a2, so that a is real or purely imaginary. If a is real then so are d and c, and we are done. If a ia0 with a0 real, then d −i/a0 and c ic0 for some c0 real. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

13. “Out of Nothing I Have Created a New Universe”—Bolyai 143

We have ia z a2z g(z) 0 0 . ic0z − i/a0 a0c0z − 1 Consider z εi, ε>0, in H2. We have a2εi a2εi(a c εi + 1) g(εi) 0 − 0 0 0 . 2 a0c0εi − 1 (a0c0ε) + 1 For ε small enough, this is not in H2. This case is not realized. Summarizing, we ﬁnd that if g maps H2 onto itself then az + b g(z) ,ad− bc , cz + d 1 with a, b, c, d real. Conversely, assume that g is of this form with real coefﬁcients. We claim that g(H2) ⊂ H2. (It is clear that g maps the real axis onto itself.) We have "(z) "(g(z)) . |cz + d|2 This is a simple computation. In fact, we have az + b (az + b)(cz¯ + d) "(g(z)) " " cz + d |cz + d|2 adz + bcz¯ adz − bcz " " |cz + d|2 |cz + d|2 "(z) . |cz + d|2 Thus, "(z) > 0 implies "(g(z)) > 0; in particular, g(H2) ⊂ H2. The theorem follows.

Corollary. A general M¨obius transformation that maps H2 onto itself can be represented by az + b a(−¯z) + b z → z → cz + d or c(−¯z) + d where a, b, c, d ∈ R. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

144 13. “Out of Nothing I Have Created a New Universe”—Bolyai

Proof. We need to consider only opposite Mobius¨ transformations. If g is opposite, the transformation z → g(−¯z) is direct and maps H2 onto itself since both g and z →−¯z do. The corollary follows. (Note that z →¯z does not work, since it maps H2 to the lower half-plane.)

Having agreed that the upper half-plane is the model of our new hyperbolic geometry in which the hyperbolic lines are Euclidean half-lines or semi-circles perpendicular to the boundary, we now see that the natural candidate for the group of transformations in this geometry is SL(2, R). The elements of SL(2, R), acting as linear fractional transformations, preserve angles but certainly do not preserve Euclidean distances. The question arises naturally: 2 Does there exist a distance function dH on H with respect to which the elements of SL(2, R) act as isometries? We seek a quantity that remains invariant under Mobius¨ transformations. The naive approach is to work out the Euclidean distance d(g(z), g(w)) of the image points g(z) and g(w) under az + b g(z) ,ad− bc , a,b,c,d,∈ . cz + d 1 R Here it is:

az + b aw + b d(g(z), g(w)) |g(z) − g(w)| − cz + d cw + d

(ad − bc)(z − w) d(z, w) . (cz + d)(cw + d) |cz + d||cw + d|

The Euclidean distance d(z, w) is divided by the “conformality factors” |cz + d| and |cw + d|. Notice, however, that these factors also occur in the expressions of "(g(z)) and "(g(w)) at the end of the proof of Theorem 9! Dividing, we obtain d(g(z), g(w))2 d(z, w)2 , "(g(z))"(g(w)) "(z)"(w)

and we see that this quotient remains invariant! It is now just a 2 matter of scaling to deﬁne the hyperbolic distance function dH : H × Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

13. “Out of Nothing I Have Created a New Universe”—Bolyai 145

H2 → R by d(z, w)2 cosh dH (z, w) 1 + . 2"(z)"(w) We use here the hyperbolic cosine function: cosh t (et + e−t)/2, t ∈ R. Note that cosh : [0, ∞) → [1, ∞) is strictly increasing, so that the formula for dH makes sense since the right-hand side is ≥ 1. The choice of cosh is not as arbitrary as it seems. As we will show shortly, it is determined by the requirement that measuring hyperbolic distance along a hyperbolic line should be additive in the sense that if z1, z2, and z3 are consecutive points on a hyperbolic line, then dH (z1,z2) + dH (z2,z3) dH (z1,z3) should hold. Summarizing, we obtain a model of hyperbolic plane geometry 2 on the upper half-plane H with hyperbolic distance function dH and group of direct isometries SL(2, R)/{±I}. (Here we inserted {±I} back for precision.) The hyperbolic lines are Euclidean half- lines and semicircles meeting the boundary of H 2 at right angles. As in Euclidean geometry, the segment on a hyperbolic line between two points is the shortest possible path joining these two points. We emphasize here that “shortest” means with respect to dH ! By the conformal (angle-preserving) nature of this model, the hyperbolic angles are ordinary Euclidean angles. Let us take a closer look at dH by traveling along a vertical line toward the boundary of H2 in Figure 13.3.

Figure 13.3 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

146 13. “Out of Nothing I Have Created a New Universe”—Bolyai

Let z a + si and w a + ti. We have (s − t)2 s2 + t2 1 s t cosh dH (z, w) 1 + + . 2st 2st 2 t s Thus,

dH (z, w) |log(s/t)|.

Note that additivity of dH along a hyperbolic line is apparent from this formula. If we ﬁx s and let t → 0, we see that d(z, w) tends to ∞ in logarithmic order. More plainly, the distances get more and more distorted as we approach the boundary of H2. Another novel feature of hyperbolic geometry is termed as “angle of parallelism.” Consider a segment l0 on a hyperbolic line which, by the abundance of hyperbolic isometries, we may assume to be a vertical line connecting z and w as above. There is a unique hyperbolic line l through w that meets l0 at a right angle. In our setting, l is nothing but the Euclidean semicircle through w with center at the vertical projection of l0 to the boundary. Let ω denote the ideal point of l to the right of l0. Finally, let l be the unique segment of the hyperbolic line (semicircle) from z to ω. We call l the right-sensed parallel to l (Figure 13.4). The angle θ at z between l0 and l is called the angle of parallelism. The length of l0 (denoted by the same letter) and θ determine each

l ′ l0

π/2 w

Figure 13.4 ω Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

13. “Out of Nothing I Have Created a New Universe”—Bolyai 147 other uniquely. Since, unlike Euclidean distances, angles are absolute in both geometries (in the sense that they have absolute unit of measurement), we ﬁnd that, through the angle of parallelism, hyperbolic distances are also absolute! Taking a closer inspection of Figure 13.4, we realize that we can actually derive an explicit formula relating θ and l0. Indeed, inserting some crucial lines, we arrive at the more detailed Figure 13.5. The Pythagorean Theorem tells us that r2 s2 + (r − t)2,z a + si, w a + ti. Using this, we ﬁnd s 2st sin θ . r s2 + t2

Comparing this with our earlier computation for dH , we see that the right-hand side is the reciprocal of cosh of (the hyperbolic length of) l0! We thus arrive at the angle of parallelism formula

sin θ cosh l0 1. Encouraged by this, we expect that there must be a formula expressing the area A of a hyperbolic triangle T in terms of its angles α, β, and γ.Hereitis: A π − (α + β + γ).

′ l0 l r π/2 w

θ Figure 13.5 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

148 13. “Out of Nothing I Have Created a New Universe”—Bolyai

π/2 β

β Figure 13.6

The right-hand side is called angular defect for obvious reasons. For the proof, we ﬁrst start with an “asymptotic triangle” with angles α π/2 and γ 0. As before, we may assume that the sides of the angle γ are vertical, as in Figure 13.6. We arrange (by an isometry) for the left vertical side to be on the imaginary axis and the ﬁnite side to be on the unit circle. Or- thogonal angles being equal, a look at Figure 13.6 shows that the right vertical side projects down to the real axis at cos β. Using the logarithmic growth formula for hyperbolic distances, we see that the element of arc length in H2 is ds |dz|/"(z). Hence, the metric is ds2 (dx2 + dy2)/y2, where z x + yi. By calculus, the area element is thus dxdy/y2. We compute the area A of T as follows: cos β ∞ dy A dx √ 2 0 1−x2 y

cos β dx √ 2 0 1 − x π/2 − β π − (π/2 + β).

Still in the asymptotic category, it is easy to generalize this to the area formula for a triangle with angles α, β arbitrary and γ 0 (we do this by pasting two asymptotic triangles together along a Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

13. “Out of Nothing I Have Created a New Universe”—Bolyai 149 vertical side). Adding areas, we arrive at A π − (α + β).

Finally, the general formula follows by cutting and pasting; that is, by considering the area of a hyperbolic triangle with nonzero angles as an algebraic sum of the areas of three asymptotic triangles. As a consequence of the angular defect formula, we ﬁnd a new non-Euclidean phenomenon: All hyperbolic triangles have area ≤ π. We derived the angular defect formula as a consequence of the special form of the hyperbolic distance in our model. In an axiomatic setting, the angular defect formula can be derived starting from Lobachevsky’s angle of parallelism, proving that all trebly asymptotic triangles (triangles with three ideal vertices) are congruent, and ﬁnally following2 the argument of Bolyai (as published in Gauss’s collected works). The angular defect formula is strikingly similar to Albert Gi- rard’s spherical excess formula,3 stating that the area of a spherical triangle in S2 is α + β + γ − π, where α, β, and γ are the spherical angles of the triangle at the vertices. A proof of this is as follows. The extensions of the sides of the spherical triangle give three great circles that divide the sphere into eight regions, the original triangle whose area we denote by A and its opposite triangle, three spherical triangles with a common side to the original triangle whose areas we denote by X, Y and Z, and their opposites. Let α, β, and γ denote the spherical angles of the original triangle opposite to the sides that are common to the other three triangles with areas X, Y, and Z. Notice that A + X is the area of a spherical wedge with spherical angle α. We thus have A + X 2α. Similarly, A + Y 2β and A + Z 2γ. Adding, we obtain 3A + X + Y + Z 2(α + β + γ). On the other hand,

2See H.S.M. Coxeter, Introduction to Geometry, Wiley, 1969. 3This appeared in A. Girard’s Invention nouvelle en alg`ebre in 1629. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

150 13. “Out of Nothing I Have Created a New Universe”—Bolyai

2(A + X + Y + Z) 4π, the area of the entire sphere. Subtracting, the spherical excess formula follows. As another consequence of the hyperbolic distance formula, we 2 now describe how a hyperbolic circle looks. Let p0 ∈ H and r>0 and work out

2 {z ∈ H | dH (z, p0) r}. We can rewrite the deﬁning equation as

cosh dH (z, p0) cosh r, or equivalently d(z, p )2 1 + 0 cosh r. 2"(z)"(p0)

Writing z x + yi and p0 a + bi, we have (x − a)2 + (y − b)2 2(cosh r − 1)by. This gives (x − a)2 + (y − b cosh r)2 (b sinh r)2, where we used the identity cosh2 r − sinh2 r 1. But this is the equation of a Euclidean circle! The radius of this circle is b sinh r, and the center has coordinates a and b cosh r. Compar- ing (a, b cosh r) with p0 (a, b), we see that p0 is closer to the boundary of H2. Thus, a hyperbolic circle looks like a Euclidean circle with center moved toward the boundary (Figure 13.7). We now go back to the roots of this long line of arguments. Remember that the starting point was our study of Mobius¨ transformations—finite compositions of rections in circles and lines. After much ado, we concluded that analytically they are given by linear fractional transformations (possibly precomposed by conjugation). Restricting to real coefficients, we understood that they act as isometries of the upper half-plane model H2 of hyperbolic geometry. In the decomposition of a Mobius¨ transformation as a finite composition of rections, we must take rections that map H2 onto itself, and these are exactly the ones with hyperbolic line axes! Thus, for a single rection, we have the following cases depicted in Figure 13.8. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

13. “Out of Nothing I Have Created a New Universe”—Bolyai 151

Figure 13.7

Let us explore some of the possible combinations. If we compose two rections with distinct vertical axes, we get an ordinary Euclidean translation along the real axis such as z → z + 1. The situation is the “same” if we consider rection to a vertical axis followed by rection in a circle with one common endpoint (Fig- ure 13.9), or composition of rections in circles with one common endpoint (Figure 13.10). Notice that parallel vertical lines also have a common endpoint at infinity. Isometries of H2 of this kind are called parabolic. If the axes intersect in H2, then the intersection is a fixed point of the composition (Figure 13.11). These isometries of H2 are called elliptic. Without loss of generality, we may assume that the fixed point is i ∈ H2. (This is because SL(2, R) carries any point to any point in H2.) Let g be an elliptic

Figure 13.8 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

152 13. “Out of Nothing I Have Created a New Universe”—Bolyai

Figure 13.9

Figure 13.10

isometry with ﬁxed point i. Since g is direct, we can write az + b g(z) ,ad− bc , a,b,c,d ∈ . cz + d 1 R Since i is a ﬁxed point, ai + b i . ci + d

Figure 13.11 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

13. “Out of Nothing I Have Created a New Universe”—Bolyai 153

Multiplying, we get a d and b −c. Thus, ad − bc a2 + b2 1. This calls for the parameterization a cos θ, b −sin θ, θ ∈ R, so that we arrive at θ · z − θ g(z) cos sin . sin θ · z + cos θ The careless reader may now say, “Well, we have not discovered anything new! After all, the matrix

cos θ − sin θ sin θ cos θ represents Euclidean rotation on R2.”But this is a grave error, since this matrix is viewed as an element of SL(2, R), and thereby represents a hyperbolic rotation in H2 given by a linear fractional transformation. If you are still doubtful, substitute θ π (not 2π!) and verify that g is the identity (and not a half-turn). Finally, isometries arising from a no-common-endpoint case are called hyperbolic. This confusing name has its own advantages. Do not confuse the term “isometry of the hyperbolic plane” with “hyperbolic isometry.” A hyperbolic isometry possesses a unique hyperbolic line, called the translation axis, perpendicular to both axes (Figure 13.12). Indeed, as the drawing on the left of Figure 13.12 shows, in case one of the axes is vertical with ideal point at c ∈ R, then the translation axis is the unique Euclidean semicircle with center at c, intersecting the other axis at the point of tangency with the radial line from c. In general (see the drawing at the right of Figure 13.12), one of the axes can be brought to a vertical Eu- clidean line by a suitable isometry, and the previous construction

Figure 13.12 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

154 13. “Out of Nothing I Have Created a New Universe”—Bolyai

applies. Clearly, the translation axis is the unique hyperbolic line that the hyperbolic isometry leaves invariant. A typical example for a hyperbolic isometry is z → kz, k 1, with the positive imaginary axis as the translation axis. In fact, all hyperbolic isometries are conjugate to z → kz for some k 1, and the conjugation is given by an isometry that carries the translation axis to the positive imaginary axis. Since in this special case the two axes of rection of the hyperbolic isometry are on concentric Euclidean circles, we have proved the following: Two nonintersecting circles can be brought to a pair of concentric circles by a Mobius¨ transformation. Indeed, if the two circles are disjoint and nonconcentric, then by a suitable Euclidean isometry that carries the line passing through their centers to the real axis, this configuration can be carried into a setting in hyperbolic geometry. As an application of these constructions, we now realize that we have solved the famous Apollonius problem: Find a circle tangent to three given circles. Indeed, first subtract the smallest radius from the three radii to reduce one circle to a point. If the remaining two circles have a common point, then this point can be carried to ∞ byaMobius¨ transformation, and the Apollonius problem reduces to finding a circle tangent to two lines and passing through a given point. If the reduced circles are disjoint, then they can be brought to two concentric circles, and once again, the Apol- lonius problem reduces to a trivial one: Find a circle tangent to a pair of concentric circles and passing through a given point. (Notice that the cases where the Apollonius problem has no solution can now be easily listed.) You may feel mildly uncomfortable with the expression “find.” To be precise, we really mean here geometric construction as explained in Problem 4 in Section 6. Now realize that all steps that involve Mobius¨ transformations are constructible. Looking back to the classification of isometries of the hyperbolic plane, we see that parabolic and hyperbolic isometries are of infinite order (that is, all their powers are distinct, or equivalently, their powers generate infinite cyclic groups). We close this section with a final note: Hyperbolic geometry has several models, and each has its own advantages and disadvan- tages. One prominent model (also due to Poincare)´ is based on the Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

13. “Out of Nothing I Have Created a New Universe”—Bolyai 155 unit disk D2 {z ∈ C ||z| < 1}. It is easy to derive this from the upper half-plane model H2. In fact, consider the linear fractional transformation i + z z → i . i − z

It is easy to see that this transformation sends D2 onto H2. (Indeed, this transformation ﬁxes ±1, sends i to ∞, and sends 0 to i.)

Remark. ♥ Since this formula seems farfetched, we elaborate on it a little. Recall from Section 11that conjugating spherical isometries of S2 by ˆ the stereographic projection hN gives Mobius¨ transformations of C. We now claim that the linear fractional transformation of D2 to H2 above is obtained from the spherical rotation Q around the real axis −1 2 with angle π/2. Indeed, hN maps D to the southern hemisphere of S2, Q rotates this to the “front” hemisphere between the (prime) ◦ ◦ 2 0 and 180 meridians of longitude, and ﬁnally, hN maps this to H . −1 Tocheck that hN ◦Q ◦hN gives the linear fractional transformation above is an easy computation.

♣ Using the linear fractional transformation of D2 to H2,we see that whatever we developed in H 2 can now be transported to D2. We ﬁnd that hyperbolic lines in D2 are segments of circles meeting the boundary S1 of D2 perpendicularly. In particular, if s is the Euclidean distance beween the origin (the center of D2) and the “midpoint” of a hyperbolic line, and r is the Euclidean radius of the circular segment representing the hyperbolic line, then perpendicularity at the boundary gives

(s + r)2 1 + r2

(Figure 13.13). Hyperbolic isometries of D2 are generated by rections in hyperbolic lines. Analytically, the isometries of D2 have the form az + c¯ az¯ + c¯ z → z → , cz + a¯ or cz¯ + a¯ Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

156 13. “Out of Nothing I Have Created a New Universe”—Bolyai

r 1

Figure 13.13

2 2 2 where |a| −|c| 1. The hyperbolic metric dD on D takes the form dD(z, w) d(z, w) tanh ,z,w∈ D2, 2 |1 − zw¯ | or equivalently, |1 − zw¯ |+|z − w| dD(z, w) log . |1 − zw¯ |−|z − w| With these distance formulas, the linear fractional transformation of D2 to H2 above becomes an isometry between the two models.

Problems

1. Show that every complex linear fractional transformation of the unit disk D2 onto itself is of the form z → eiθ(z − w)/(1 − zw)¯ , where w ∈ D2 and θ ∈ R (cf. Problem 4 of Section 7). 2. Solve Monge’s problem: Find a circle orthogonal to three given circles.

2 3. In the upper half-plane H , consider two Euclidean rays l1 and l2 emanating from a boundary point ω. (a) Show that all circular segments l0 with center ω connecting l1 and l2 have the same hyperbolic length. (Hint: Use hyperbolic rections in semi-circles with center at ω.) (b) What is the connection between the angle of parallelism for the common hyperbolic length of l0 in (a) and the angle between l1 and l2 at ω? 4. Consider a hyperbolic right triangle with hyperbolic side lengths a, b, and hypotenuse c, and angles α, β, and π/2. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 157

(a) Prove the Pythagorean Theorem: cosh a cosh b cosh c. (b) Show that sinh a tan β tanh b. 5. Cut a sphere of radius 1/2 into two hemispheres along the equator and keep the southern hemisphere H. Let H sit on R2 with the South Pole touching the plane. Let h : H → D2 be the stereographic projection from the North Pole N ( , , ) v H → D2 D2 0 0 1 , and let : 1/2 be the vertical projection to the disk 1/2 / v ◦ h−1 D2 → D2 with center at the origin and radius 1 2. Show that : 1/2 maps D2 D2 hyperbolic lines in to chords in 1/2. Using this correspondence, develop D2 hyperbolic geometry in 1/2. (This model is due to F. Klein.) 6. In the upper half-space model H3 {(z, s) ∈ C × R | s>0}⊂C × R ∪ {∞} of hyperbolic space geometry, hyperbolic lines are Euclidean semicircles or half-lines meeting the ideal boundary Cˆ {(z, 0) | z ∈ C} ∪ {∞}

at right angles. The hyperbolic distance function dH : H3 ×H3 → R is defined by |z − w|2 +|s − t|2 3 cosh dH ((z, s), (w, t)) 1 + , (z, s), (w, t) ∈ H . 2st ˆ Given a rection RS in a circle S Sr (p0) on the ideal boundary C, define the Poincaré extension R˜ S as rection to the upper hemisphere (in H3) with boundary circle S. Given a Mobius¨ transformation g of the ideal boundary Cˆ , define the Poincaré extension g˜ : H3 → H3 by decomposing g into a finite composition of rections and taking the Poincar e´ extension of each factor. (a) Show that g˜ is well defined (that is, it does not depend on the particular representation of g as a composition of rections). (b) Verify that g˜ is a hyperbolic isometry of H3. (The Poincare´ extension thus defines Möb+(Cˆ ) as a subgroup of Iso (H3).) 7. Develop hyperbolic space geometry in the 3-dimensional unit ball D3 {p ∈ R3 ||p| < 1}

with hyperbolic distance function dD : D3 × D3 → R deﬁned by |p − q|2 2 3 cosh(dD(p, q)) 1 + ,p,q∈ D . (1 −|p|2)(1 −|q|2) Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Fuchsian Groups . . . 14 . SECTION .

♥ We are now ready to discuss the final task in hyperbolic geometry: classification of discrete subgroups of the group of direct hyperbolic isometries in SL(2, R). Unlike the case of crystallographic groups, we do not consider opposite Mobius¨ transformations here. However, this does not prevent us from con- structing these discrete subgroups by means of opposite isometries (such as groups generated by rections in the sides of a hyperbolic triangle; see the discussion of triangle groups at the end of this section). Here we are entering an area in which full understanding has not been obtained, although a vast number of results are known. To be modest, we will only give a few illustrations that reveal some subtleties of the subject. First some definitions. Given a linear fractional transformation az + b g(z) ,ad− bc , a,b,c,d ∈ , cz + d 1 C we introduce its norm: |g| |a|2 +|b|2 +|c|2 +|d|2.

This is well deﬁned since the right-hand side is unchanged if we replace a, b, c, d by their negatives. (Remember the ambiguity 158 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

14. Fuchsian Groups 159

+ {±I}.) A subgroup G ⊂ M¨ob (Cˆ ) is discrete if for any n>0, the set

{g ∈ G ||g|

G(z) {g(z) | g ∈ G},z∈ H2,

2 exactly once. A fundamental domain for G is a domain F0 in H ¯ such that there is a fundamental set F for G between F0 and F0 and ¯ 2 ∂F0 F0 − F0 has zero dimensional area. The simplest Fuchsian groups, G, are cyclic; that is, they are generated by a single linear fractional transformation g ∈ SL(2, R). We write this as G 'g(. Depending on whether g is parabolic, hyperbolic, or elliptic, we arrive at the following examples:

1The terms ‘Kleinian’ and ‘Fuchsian’ are due to Poincare.´ 2A word of caution. The closure here is taken in H2, and by “area” we mean hyperbolic area. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

160 14. Fuchsian Groups

Figure 14.1

Example 1 Let g be the parabolic isometry

g(z) z + 1,z∈ H2.

A fundamental domain F0 for G 'g( is given by 0 < !(z) < 1, shown in Figure 14.1. Figure 14.2 shows the transformed fundamental domain and tiling on D2.

Figure 14.2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

14. Fuchsian Groups 161

Remark. It is instructive to look at the transformation g from D2 to H2 via −1 hN ◦ Q ◦ hN , where Q is a quarter-turn around the real axis (cf. Section 13).

Example 2 Let k>1 and consider the hyperbolic isometry g(z) kz, z ∈ H2.

A fundamental domain for G 'g( is given by 2 F0 {z ∈ H | 1 < |z|

Example 3 Let g be elliptic with ﬁxed point at i ∈ H2. As noted above, g can be written as cos θ · z − sin θ g(z) ,z∈ H2, sin θ · z + cos θ and discreteness implies that θ π/n for some n ≥ 2. The tiling for G 'g( is shown in Figure 14.5 for H2 and in Figure 14.6 for D2 (noting that i corresponds to the origin).

Before we leave the cyclic Fuchsian groups, here is one more simple example, in which G is generated by an elliptic and a hyperbolic isometry.

Figure 14.3 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

162 14. Fuchsian Groups

Figure 14.4

Figure 14.5

Figure 14.6 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

14. Fuchsian Groups 163

Figure 14.7 Example 4

Let k>1 and G 'g1,g2(, where 2 g1(z) −1/z and g2(z) kz, z ∈ H . A fundamental domain and the tiling is shown in Figure 14.7.

Examples 1 to 4 (and their conjugates in SL(2, R))makeupwhat we call elementary Fuchsian groups. From now on we concentrate on nonelementary Fuchsian groups.

Example 5

Let G SL(2, Z) be the modular group. We claim that F0 deﬁned by |z| > 1 and |!(z)| < 1/2 is a fundamental domain (Figure 14.8). Let g be an arbitrary element in G and write az + b g(z) ,ad− bc , a,b,c,d ∈ . cz + d 1 Z

Assume that z ∈ F0, i.e., the deﬁning inequalities for F0 hold. We compute |cz + d|2 c2|z|2 + 2!(z)cd + d2 >c2 + d2 −|cd| (|c|−|d|)2 +|cd|. The lower bound here is a nonnegative integer and is zero iff c d 0. This cannot happen since ad − bc 1. Thus, 2 |cz + d| > 1,z∈ F0. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

164 14. Fuchsian Groups

Figure 14.8

Using this, we have "(z) "(g(z)) < "(z), z ∈ F , |cz + d|2 0 where we used the formula for "(g(z)) in the proof of Theorem 9. We now show that F0 contains at most one point from each orbit of G. ¬ Assuming the contrary, we can ﬁnd z ∈ F0 and g ∈ G such that g(z) ∈ F0. By the computation above, "(g(z)) < "(z).

Replacing z by g(z) and g by g−1 in this argument, we obtain "(g−1(g(z))) "(z) < "(g(z)),

and this contradicts the inequality we just obtained! ¬ The parabolic isometry z → z + 1 and the elliptic “half-turn” z →−1/z are in G, and by applying them, we can easily move 2 ¯ any point in H to F0. Thus, F0 is a fundamental domain for G. (As a by-product, we obtain that SL(2, Z) is generated by z → z + 1 and 2 z →−1/z, z ∈ H .) Notice that F0 is a hyperbolic triangle (in the asymptotic sense with one vertex at inﬁnity). A fundamental set F ¯ with F0 ⊂ F ⊂ F is given by the inequalities −1/2 < !(z) ≤ 1/2, |z|≥1, and !(z) ≥ 0if|z| 1. Geometrically, we add to F0 the right half of the boundary circle and the right vertical side of F0. Thus, the modular group tiles H2 with triangles (Figure 14.9). Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

14. Fuchsian Groups 165

Figure 14.9

Remark. At the end of Section 2, we introduced the modular group SL(2, Z) by studying various bases in a lattice in R2 C2 (cf. Problem 15). Since we constructed a fundamental set F for SL(2, Z) above, it follows that for all the bases {v, w} of a given lattice L ⊂ C there is a unique ratio τ w/v in F. A basis {v, w} with this property is called canonical. Given a ratio τ ∈ F corresponding to a canonical basis, there is a choice of two, four, or six canonical bases with this ratio. Indeed, {−v, −w} is always canonical, and more canonical bases occur if τ is the ﬁxed point of an element in SL(2, Z). This happens only if τ i (z →−1/z) and τ ω e2πi/3 (z → −(z + 1)/z, −1/(z + 1)).

The complexity of Fuchsian groups increases as we look at more and more “random” examples such as the following:

Example 6

Let G 'g1,g2(, where 3z + 4 g (z) and g (z) 2z, z ∈ H2. 1 2z + 3 2 A fundamental domain for G is shown in Figure 14.10.

−2 −121 Figure 14.10 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

166 14. Fuchsian Groups

A group G of isometries of H2 is said to be a triangle group of type (α, β, γ)ifG is generated by rections in the sides of a hyperbolic triangle with angles α, β, and γ. Here α, β, γ ≥ 0, and we have α + β + γ<π.

Notice that G cannot be Fuchsian, since it contains opposite isometries. We remedy this by considering the subgroup G+ of direct isometries in G. The elements of G+ are compositions of elements in G with an even number of factors. We call G+ a conformal group of type (α, β, γ). For example, if l, l, and l denote the sides of a hyperbolic triangle, then Rl ◦ Rl, Rl ◦ Rl (and Rl ◦ Rl ) all belong to (in fact, generate!) G+. Let the triangle be as shown in Figure 14.11. Rl ◦ Rl fixes p so that it is either elliptic (with rotation angle 2α, α>0) or parabolic (α 0). Similarly, Rl ◦ Rl fixes q and is elliptic or parabolic depending on whether β>0orβ 0. Thus + G is generated by two isometries g1, g2, each being parabolic or elliptic. If G+ is discrete, then every elliptic element must have finite order in G+. It follows that if α, β, and γ are positive, then π/α, π/β, and π/γ are rational.

Example 7 Let T be the asymptotic triangle in H2 with vertices z(π/3), ∞, and i as shown in Figure 14.12. The triangle group G corresponding to T

α l ′ l

β γ ″ q l r

Figure 14.11 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

14. Fuchsian Groups 167

π/2 2π/3

Figure 14.12

is of type (π/3, 0,π/2). Hyperbolic rections in the sides of T give z →−¯z, z → 1/z¯, and z →−¯z +1, z ∈ H2. Takingcompositions of these, we see that the corresponding conformal group G+ of type (π/3, 0,π/2) is generated by the translation z → z + 1 and the half-turn z →−1/z, z ∈ H2. In perfect analogy with Example 5, we obtain that G+ is the modular group SL(2, Z).

Poincare´ proved that a triangle group of type (α,β,γ)is discrete if π/α, π/β, and π/γ are integers ≥ 3 (possibly ∞) with

α β γ + + < π π π 1

(cf. the remark at the end of Section 10). It can also be proved that the corresponding conformal group of type (α,β,γ)tiles the hyperbolic plane H2. Thus, the situation is radically different from the Euclidean case of crystallographic groups, where there are only ﬁnitely many (in fact, 17) different tilings. You are now invited Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

168 14. Fuchsian Groups

Figure 14.13

to look at Figure 14.13, depicting a hyperbolic tiling of a triangle group3 of type (π/6,π/4,π/2), and discover various elliptic isometries of the corresponding conformal group. Our last two examples involve regular n-sided hyperbolic polygons P in D2 with centroid at the origin. Let c be the center of a circle that contains a side of P, m the midpoint of the side, and v a vertex of P adjacent to m. Let r be the Euclidean radius of the circle and s the Euclidean distance between the origin 0 and m. Figure 14.14 depicts the case n 3. Finally, let α 0vm. Since v0m π/n, the (Euclidean) law of sines gives s + r r , cos α sin(π/n) where we used the identity sin(α + π/2) cos α. As noted at the end of Section 13, we have (s + r)2 1 + r2.

3See H.S.M. Coxeter, Introduction to Geometry, Copyright 1969 by John Wiley & Sons, Inc. Reprinted by permission of John Wiley & Sons, Inc. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

14. Fuchsian Groups 169

r 1 v π/24 s mc

Figure 14.14

Combining these, we obtain (π/n) r sin cos2 α − sin2(π/n) and α − (π/n) s cos sin . cos2 α − sin2(π/n)

Example 8 A triangle group G of type (π/4,π/4,π/4) is generated by rections in the sides of a hyperbolic (equilateral) triangle with α π/8 as shown in Figure 14.14. The formulas above reduce to √ √ √ 3 2 + 2 − 3 r √ and s √ . 2 − 1 2 − 1 With these, we have v s + r(1 + z(23π/24)). The corresponding conformal group G+ of type (π/4,π/4,π/4) is generated by elliptic quarter-turns around the vertices v and z(2π/3)v. What is a suitable fundamental domain for G+?

Example 9 Finally, we discuss tilings of H2 with fundamental polygons of 4p sides, p ≥ 2, and Fuchsian groups generated by 2p hyperbolic elements. We consider in full detail the case p 2 of a Fuchsian group Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

170 14. Fuchsian Groups

(s+r)i

v α si

Figure 14.15

generated by four hyperbolic elements. A fundamental polygon is the hyperbolic octagon depicted in Figure 14.15. Each hyperbolic isometry will map a side of the octagon to the opposite side. We ﬁrst work out the hyperbolic isometry g that maps the bottom side of the octagon to the top side. Let si ∈ D2 be the midpoint of the top side, and v an adjacent vertex of the octagon. The hyperbolic triangle with vertices 0, si and v has angle α π/8at0andv. This is because D2 is tesselated by images of the octagon under the Fuchsian group. Extending a side to a circle of radius r, the formulas above reduce to sin α cos α − sin α r and s . cos(2α) cos(2α) The hyperbolic distance of si from the origin computes as 1 + s cos α + cos(2α) dD(si, 0) log log 1 − s sin α (see Section 13). As in Example 2, we represent g on H2 by z → kz, z ∈ H2, and the normalization in SL(2, R) implies that the matrix associated to g is √ k 0√ . 01/ k Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 171

On D2, g maps −si, the midpoint of the bottom side, to si, the midpoint of the top side. By symmetry, the hyperbolic distance 2 between these points is 2dD(si, 0)n H , the hyperbolic distance between i and ki is log k (see Section 13). It follows that

log k 2dD(si, 0), so that √ cos α + cos(2α) k . sin α The formula representing g as a linear fractional transformation follows. By using symmetries of the octagon, we see that g, conjugated by rotations around the origin with angles π/4, π/2, and 3π/4, deﬁnes the three other hyperbolic isometries. Replacing α π/8byα π/4p, p ≥ 2, we obtain a Fuch- sian group generated by 2p hyperbolic elements with fundamental domain a regular 4p-sided hyperbolic polygon.

Problems

1. Show that an isometry of the hyperbolic plane H2 is conjugate in SL(2, Z) to (a) z → z + 1 iff it is parabolic; (b) to z → (cos θ · z − sin θ)/(sin θ · z + cos θ), θ ∈ R, iff it is elliptic; (c) z → k · z, k 1, iff it is hyperbolic. 2. Characterize the parabolic, elliptic, and hyperbolic isometries of the hyperbolic plane in terms of their fixed points on H2 ∪ R ∪∞. 3. Let g(z) (az + b)/(cz + d), ad − bc 1, a, b, c, d ∈ R, be an isometry of H2. Show that trace2(g) (a + d)2 is well defined (that is, it depends only on g and not on a, b, c, d). Prove that g is parabolic iff trace2(g) 4, elliptic iff 0 ≤ trace2(g) < 4 and hyperbolic iff trace2(g) > 4. (The latter two relations imply that trace2(g) is real.) 4. Use Problems 2–3 to conclude that an isometry g of H2 has no fixed point in H2 iff trace2(g) ≥ 4. 5. Let g be a linear fractional transformation that satisfies gn(z) z for some n ≥ 2. Show that g is elliptic.

2 6. Let g1 and g2 be isometries of H such that g1 ◦ g2 g2 ◦ g1. Show that g1 parabolic implies that g2 is also parabolic. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

172 14. Fuchsian Groups

2 7. Let T1 and T2 be adjacent asymptotic hyperbolic triangles in H with vertices z(π/3), ∞, i and z(π/3),1,∞ (see Example 7). Show that the conformal groups corresponding to T1 and T2 are both equal to SL(2, Z). (Thus, incongruent triangles can deﬁne the same conformal group.) 8. Work out the vertices of a regular hyperbolic pentagon in D2 (with centroid at the origin) whose consecutive sides are perpendicular. Generalize this to regular n-sided polygons in D2 for n ≥ 5. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Riemann Surfaces . . . 15 . SECTION .

♦ Having completed our long journey from Euclidean plane geometry through crystallographic groups, Mobius¨ transformations, and hyperbolic plane geometry, we are now ready to gain a geometric insight into the structure of Fuchsian groups on H2. Before this, however, we need a bit of complex calculus. Just like linear fractional transformations, complex functions are usually deﬁned on open sets U of the complex plane C and have their values in C. Given a complex valued function f : U → C, we say that f is differentiable (in the complex sense)atz0 ∈ U if the limit f(z) − f(z ) lim 0 z→z 0 z − z0

exists. In this case, we call the limit the derivative of f at z0 and denote it by f (z0). We say that f is differentiable (in the complex sense)onU if f (z0) exists for all z0 ∈ U. One of the most stunning novelties of complex calculus is that the existence of the ﬁrst derivative of f on U implies the existence of all higher derivatives f (n), n ∈ N,onU! (This is clearly nonsense in real calculus; for example, f : R → R deﬁned by xn, x ≥ f(x) if 0 −xn, if x<0 173 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

174 15. Riemann Surfaces

has derivatives up to order n − 1atx0 0, but n , x ≥ f (n)(x) ! if 0 −n!, if x<0

is discontinuous at x0 0.) As a matter of fact, even more is true! If f exists on U, then f is analytic on U; that is, given z0 ∈ U, f can be expanded into a power series ∞ n f(z) cn(z − z0) n 0

that is absolutely convergent in a neighborhood of z0 (contained in U). (♠ Like almost everything in complex calculus, this is a consequence of the Cauchy formula.) ♦ The coefﬁcients of the expansion have no other choice but to be equal to (n) f (z0) cn , n! so that the power series is actually Taylor. (Once analyticity is accepted, this follows since an absolutely convergent series can be differentiated term by term.) From now on we use the term analytic for a complex function f : U → C whose derivative exists on U. The usual rules of differentiation are valid, and the proofs are the same as in the real case. In particular, every rational function (such as a linear fractional transformation) a + a z +···+a zn 0 1 n , m b0 + b1z +···+bmz

an 0 bm,a0,...,an; b0,...,bm ∈ C, is analytic everywhere except at ﬁnitely many (≤ m) points where the denominator vanishes. Moreover, and this is very important for Riemann surfaces, composition of analytic functions is analytic. Complex functions f : U → C are often thought of as being locally deﬁned transformations of the complex plane. As such, the existence of the derivative of f must carry a geometric meaning. Let us eleborate on this a little. Let f be analytic on U and assume that f (z0) 0atz0 ∈ U. Let γ : (−a, a) → C, a>0, be a Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

15. Riemann Surfaces 175 smooth curve through z0 with γ(0) z0. Consider the image f ◦ γ. Differentiating, we get (f ◦ γ) (0) f (γ(0)) · γ (0) f (z0)γ (0). This means that |(f ◦ γ) (0)| |f (z0)|·|γ (0)| and arg(f ◦ γ) (0) arg f (z0) + arg γ (0).

Thus, f acts on tangent vectors at z0 by multiplying them by the constant |f (z0)| and rotating them by arg f (z0). In particular, f preserves (signed) angles, a property that we note by saying that f is conformal. Summarizing, we obtain that an analytic function is conformal1 where its derivative does not vanish! Notice that this argument applies to linear fractional transformations and was given earlier for Mobius¨ transformations. We begin to suspect that the big leap from ﬁrst-order differentiability to analyticity will rule out a lot of (otherwise nice) functions. The simplest example is conjugation z →¯z, z ∈ C.We claim that conjugation is nowhere differentiable. In fact, setting z − z0 r(cos θ + i sin θ) and keeping θ ﬁxed, we have z¯ −¯z r(cos θ − i sin θ) lim 0 lim z→z r→ 0 z − z0 0 r(cos θ + i sin θ) cos θ − i sin θ cos(2θ) − i sin(2θ), cos θ + i sin θ and for different arguments θ, this takes different values! Thus, the limit and hence the derivative do not exist. Using this, a number of nondifferentiable complex functions can be manufactured; f(z) 1/z¯ will appear shortly. (♠ You know, of course, the underlying theme that differentiability in complex sense is smoothness plus the Cauchy-Riemann equations. These we avoided as part of our desperate effort to keep the length and level of the exposition to a minimum.)

1The connection between complex differentiability and conformality was ﬁrst recognized by Gauss in 1825. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

176 15. Riemann Surfaces

♦ Let f : U → C be analytic on U and f (z0) 0atz0 ∈ U. The Taylor expansion of f at z0 implies that m f(z) − f(z0) (z − z0) g(z) for some m ≥ 2, where g is analytic and nonzero on an open neighborhood U0 ⊂ U of z0. Choosing a suitable branch of the mth root of g, we have m m f(z) − f(z0) (z − z0) g(z) .

In other words, f is the mth power of an analytic function h deﬁned on U0 by m h(z) (z − z0) g(z), z ∈ U0.

Notice that h(z0) 0 and h (z0) 0, so that h establishes a conformal equivalence between a neighborhood of z0 (in U0) and a neighborhood of the origin. It is now time to introduce the most basic elementary function in complex calculus: the exponential function. We deﬁne it by the power series ∞ zn ez , n n 0 ! which is clearly convergent on the entire complex plane. In particular, (ez) ez. When z r is real, this reduces to the ordinary exponential function. To see what happens in the imaginary direction, we take z iθ, θ ∈ R. We compute ∞ inθn eiθ n n 0 ! ∞ i2kθ2k ∞ i2k+1θ2k+1 + ( k) ( k + ) k 0 2 ! k 0 2 1 ! ∞ (− )kθ2k ∞ (− )kθ2k+1 1 + i 1 ( k) ( k + ) k 0 2 ! k 0 2 1 ! cos θ + i sin θ. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

15. Riemann Surfaces 177

Notice that in the second equality we split the infinite sum into two sums; one running on even indices (n 2k), the other on odd indices (n 2k+1). This is legitimate, since the series is absolutely convergent. Then we used the fact that i2k (i2)k (−1)k, and finally we remembered the Taylor expansion of sine and cosine. We arrive at the famous Euler formula eiθ cos θ + i sin θ, θ ∈ R. This is our old friend z(θ), whom we met a long time ago when discussing complex arithmetic! Multiplying through by r, we obtain the exponential form of a complex number: z reiθ, |z| r, arg z θ + 2kπ, k ∈ Z. θ π in the Euler formula gives the equation eiπ −1, connecting the three most prominent numbers π, e, and i of mathematics. This appears in Euler’s Introductio, published in Lausanne in 1748. Aside from its commercial value shown on the T-shirts of mathematics students, “We are all number −eiπ!”, its significance can hardly be underrated. Without much explanation, we humbly recite the words of Benjamin Pierce (1809–1880) to his students: Gentlemen, that is surely true, it is absolutely paradoxical; we cannot understand it, and we don’t know what it means, but we have proved it, and therefore we know it must be the truth. The natural extension of the relation x n ex lim 1 + n→∞ n to complex exponents can easily be understood. For purely imaginary exponents Euler’s formula implies θ n eiθ lim 1 + i . n→∞ n Indeed, the argument of the complex number 1+iθ/n is tan−1(θ/n), −1 so that the argument of the right-hand side is limn→∞ n tan (θ/n) θ. The absolute value of the right-hand side is unity, since Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

178 15. Riemann Surfaces

Figure 15.1

2 2 n/2 limn→∞(1 + θ /n ) 1 (a simple consequence of the binomial formula). There is a beautiful geometric interpretation of this. For ﬁxed n, the points (1 + iθ/n)k, k 0,...,n, are vertices of n right triangles arranged in a fanlike pattern as shown in Figure 15.1 for n 10 and n 50. Complex√ multiplication by (1 + iθ/n) amounts to multiplication by 1 + θ2/n2 and rotation by the angle tan−1(θ/n). Thus, the hypotenuse of each triangle in the fan is the base of the next.

Remark. The equation 1 + eiπ 0 was used by Lindemann to prove transcendentality of π (cf. Section 4). In fact, he proved2 that in an equation of the form

a1 an c0 + c1e +···+cne 0,

the coefﬁcients c0,...cn and the exponents a1,...,an cannot all be complex algebraic numbers.

The exponential function satisﬁes the usual exponential identity

z1+z2 z1 z2 e e · e ,z1,z2 ∈ C, which follows easily from trigonometric identities. (If you try to prove this from the Taylorseries deﬁnition, you have to work a little harder and use the binomial formula.) In particular, ez is periodic with period 2πi: ez+2kπi ez,k∈ Z. The exponential map exp : C → C has image C −{0}, as can easily be seen by writing an image point in exponential form. In fact,

2For a lively account, see F. Klein. Famous Problems of Elementary Geometry, Chelsea, New York, 1955. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

15. Riemann Surfaces 179 given 0 w ∈ C, the complex numbers log |w|+i arg w (arg w has infinitely many values!) all map under exp to w. This is thus the inverse of exp, legitimately called the complex logarithm log w of w. It is now the multiple valuedness of log that gives rise to Riemann surfaces. In fact, attempting to make log single valued is the same as trying to make the exponential map one-to-one. This is what we will explain next. Consider the complex plane C . The periodicity formula for exp tells us that z and z + 2kπi, k ∈ Z, are mapped to the same point by exp. Thus, we should not consider these two points different in C! Identifying them means rolling C into a cylinder in the imaginary direction. (This can be demonstrated easily by pouring paint on the Chinese rug at home and rolling it out.) We obtain that exp actually maps the cylinder to the punctured plane (see Color Plate 2a). The rulings of the cylinder are mapped to rays emanating from the origin. (This is because the rulings can be parametrized by t → t + θi with θ ∈ R fixed, and the images are parametrized by t → et · eiθ.) Similarly, circles on the cylinder map to concentric circles around the origin. We now see that the exponential map establishes a one-to-one correspondence between the cylinder and the punctured plane. We will say later that the cylinder and the punctured complex plane are conformally equivalent Riemann surfaces. We now notice that the set of points z + 2kπi ∈ C,k∈ Z, that are to be identified form the orbit3 of the first frieze group generated by the translation T2πi! It is clear how to generalize this to obtain more subtle Riemann surfaces. We consider discrete groups on C R2 or Fuchsian groups on H2, and in each case, we identify points that are on the same orbit of the acting group. The concept of fundamental set (domain), which we used to visualize wallpaper patterns, now gains primary importance! By its very definition, it contains exactly

3For group actions, refer to the end of “Groups” in Appendix B. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

180 15. Riemann Surfaces

one point from each orbit, so that as a point-set, it is in one-to-one correspondence with the Riemann surface. In our cases, the fundamental domain is a Euclidean or hyperbolic polygon, so that when taking the closure of our fundamental set, the points that are on the same orbit (and thereby are to be identified) appear on the sides of this polygon. Thus, to obtain our Riemann surface topologically, we need to paste the polygon’s sides together in a certain manner as prescribed by the acting group. This prescription is called side-pairing transformation. Enough of these generalities. Let us now consider crystallographic groups on C R2. We first consider translation groups. Any such crystallographic group G is generated by two translations Tv and Tw with v and w linearly independent. A fundamental domain F0 for G is a parallelogram with vertices 0, v, w, v + w (Figure 15.2). The boundary ∂F0 consists of the four sides of the parallelogram. The restriction of the action of the group G to these sides gives the side-pairing transformations. It is clear that under Tw, tv,0 ≤ t ≤ 1, gets identified with Tw(tv) tv + w (Figure 15.3). Similarly, under Tv, tw,0≤ t ≤ 1, gets identified with Tv(tw) tw + v (Figure 15.4). Thus, the Riemann surface is obtained by pasting together the base and top sides and the left and right sides. We obtain what is called a complex torus (Figure 15.5).

Figure 15.2 v Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

15. Riemann Surfaces 181

v Figure 15.3

v Figure 15.4

Figure 15.5 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

182 15. Riemann Surfaces

You might say that all these tori (the plural of torus) look alike, so why did not we just take v and w to be orthogonal unit vectors spanning the integer lattice Z2 in R2? The answer depends on what we mean by “alike”. A topologist would certainly consider all of them the same, since they are actually homeomorphic.4 But from the analyst’s point of view, they may be different, since the conformal structure (the way we determine angles) should depend on the vectors v and w. This we will make more precise shortly. Looking back, we see that we used the term Riemann surface a number of times without actually deﬁning what it is. We now have enough intuition to ﬁll the gap properly. To tell you the truth, in the proof of the FTA we came very close to this concept! Recall the 2 2 stereographic projections hN : S −{N}→C and hS : S −{S}→C and their connecting relation

−1 N ◦ hS )(z) 1/z,¯ 0 z ∈ C.

Putting hN and hS on symmetric footing, in the proof of the FTA they were used to “view” the map f : S2 → S2 in two different ways. One was the original polynomial

−1 P hN ◦ f ◦ hN : C → C, and the other,

−1 Q hS ◦ f ◦ hS : C → C,

was a rational function. hN and hS both have the “defect” that their domains do not cover the entire sphere, but the union of these domains is S2, and so P and Q describe f completely. We will call 2 hN and hS coordinate charts for S . There is, however, one minor technical difficulty. To define the notion of analyticity of functions defined on open sets of S2,we would use hN and hS to pull the functions down to open sets of C and verify analyticity there. Take, for example, the function hN .To −1 pull this down to C we use hN itself and obtain hN ◦ hN : C → C, −1 the identity. If we use hS, however, we obtain hN ◦ hS : C → C, and this is not analytic, since z →¯z is nowhere differentiable! The

4See “Topology” in Appendix C. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

15. Riemann Surfaces 183 problem, of course, is the presence of conjugation, and it is easily ¯ remedied by taking hS instead of hS. This gives 1 ◦ h¯ −1)(z) , z ∈ , N S z 0 C and now analyticity of a function on S2 no longer depends on ¯ ¯ whether it is viewed by hN or hS! (Replacing hS with hS is natural, since z → 1/z is the simplest direct Mobius¨ transformation that is not an isometry. Notice also the role of this transformation in the proof of Theorem 8 of Section 12.) Now the general deﬁnition: ♠ A connected (Hausdorff) topological space5 M is a Riemann surface if M is equipped with a family

{ϕj : Uj → C | j ∈ N} called the atlas (each ϕj : Uj → C is called a coordinate chart of M) such that

1. ∪j∈NUj M; 2. Each ϕj is a homeomorphism of Uj to an open set of the complex plane C; 3. If U Uk ∩ Uj is nonempty, then

−1 ϕk ◦ ϕj : ϕj(U) → ϕk(U) is an analytic map between open sets of the complex plane C (Figure 15.6).

Uj Uk

φ j φ k

Figure 15.6

5See “Topology” in Appendix C. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

184 15. Riemann Surfaces

Remark. We can use a subset of the set of positive integers as the index set for the atlas, which is not required in the usual deﬁnition of a Riemann surface. It is, however, a deeper result that every Riemann surface carries a countable atlas, so that this choice can always be made. Looking back, we see that

2 ¯ 2 {hN : S −{N}→C, hS : S −{S}→C}

is an atlas for S2, so that S2 is a Riemann surface.

Analyticity is a local property, so that any open subset of a Rie- mann surface is also a Riemann surface. Thus, the complex plane C, the punctured complex plane C −{0}, etc. are Riemann surfaces. The Riemann surface structure on the cylinder and tori are derived from the Riemann surface structure of the complex plane C in the following way: Recall that they are defined by identifying the points that are on the same orbit of the acting (translation) group G. The identification space, that is, the space of orbits C/G, is a topological space under the quotient topology. Actually, the topology on C/G is defined to make continuous the natural projection π : C → C/G associating to z ∈ C its orbit G(z) {g(z) | g ∈ G}. Now let z ∈ D (z ) {z ∈ ||z−z |

is the restriction of an element in G (a translation) and thereby analytic. The cylinder and all complex tori thus become Riemann surfaces. We may still feel uneasy about the initial choice of the group G acting on C, since it consists of translations only. As a matter of fact, we may think that less trivial choices of G may lead to more subtle Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

15. Riemann Surfaces 185

Riemann surfaces C/G. That this is not the case for the complex plane C is one result in the theory of Riemann surfaces. In seeking new domains, we may also consider the extended complex plane Cˆ or, what is the same, the sphere S2. The question is the same: “Does there exist a discrete group G acting on S2 that gives a Riemann surface S2/G?” As far as the group G is concerned, the answer is certainly yes; we just have to remember the proof of the FTA, where the map z → zn induced an n-fold wrap of S2 to itself, leaving the North and South Poles ﬁxed. It is not hard to see that this map can be thought of as the projection π : S2 → S2/G, where G is generated by the rotation R2π/n around the origin in C ⊂ Cˆ . As far as the Riemann surface S2/G is concerned, it is again a result in the theory of Riemann surfaces that we do not get anything other than S2. Summarizing, the only Riemann surfaces that are quotients of S2 or C are the sphere, the complex plane, the cylinder, and the complex tori. We now take the general approach a little further and deﬁne analyticity of a map f : M → N between Riemann surfaces M and N. Given p0 ∈ M, we say that f is (complex) differentiable at p0 if the composition

−1 ψk ◦ f ◦ ϕj shown in Figure 15.7 is (complex) differentiable at ϕj(p0). Here ϕj : Uj → C is a chart covering p0; that is, p0 ∈ Uj, and ψk : Vk → C is a chart covering f(p0), so that the composition is defined near ϕj(p0). (To be perfectly precise, the composition is defined on −1 ϕj(Uj ∩ f k)), but this is really too much distraction.) We also see that we defined the concept of Riemann surfaces in just such a

f UjkV

φ ψ j k

Figure 15.7 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

186 15. Riemann Surfaces

way as to make this definition independent of the choice of charts! (You are invited to choose other charts, work out the compositions, and verify independence. Advice: Draw a detailed picture rather than engage in gory computational details.) If f : M → N is differentiable at each point of M, we say that f is analytic. As an example, it is clear from the way we defined C/G for G a discrete translation group on C, that the projection π : C → C/G is analytic. At the other extreme, an invertible analytic map (whose inverse is also analytic) is called a conformal equivalence. (The name clearly comes from the fact that if f is invertible, then its derivative is everywhere nonzero, so that f viewed as a transformation is conformal.) Finally, two Riemann surfaces are called conformally equivalent if there is a conformal equivalence between them. Our prominent example: The cylinder and the punctured complex plane are conformally equivalent Riemann surfaces. If f : M → N is an analytic map between Riemann surfaces and p0 ∈ M, then the nonvanishing of the derivative of the local −1 representation ψk ◦ f ◦ φj at p0 is independent of the choice of the charts φj and ψk. In this case, f is a local conformal equivalence between some open neighborhoods U0 of p0 and V0 of f(p0).Ifthe −1 derivative of ψk ◦f ◦φj vanishes at p0, then as the Taylorexpansion shows (cf. the argument above for a complex function f ), there are −1 local charts φj and ψk such that ψk ◦ f ◦ φj is the mth power function for some m. This is exactly the case when f has a branch point at p0 with branch number m − 1. (In fact, this can be taken as the definition of the branch point.) We now see that we used this (somewhat) intuitively in the proof of the FTA in Section 8! Liouville’s theorem in complex calculus implies that the entire complex plane C is not conformally equivalent to D2. (Another proof is based on Schwarz’s lemma, which asserts that the conformal self-maps of D2 are linear fractional transformations, and thereby they have the form given in Problem 1 of Section 13. They can be parametrized by three real parameters: !(w), "(w), and θ. In contrast, the linear transformations z → az + b, a, b ∈ C, form a 4-parameter family of conformal self-maps of the complex plane.) On the other hand, the linear fractional transformation z → i(i + z)/(i − z) restricted to the unit disk D2 establishes a con- Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

15. Riemann Surfaces 187 formal equivalence between D2 and the upper half-plane H2. More generally, the Riemann mapping theorem states that any simply connected6 domain in C that is not the whole plane is conformally equivalent to D2. The question about the tori being “alike” can now be reformulated rigorously: “Which tori are conformally equivalent?” This is a question of Riemann moduli, an advanced topic.7 A beautiful result of complex function theory asserts that two T /G T /G G 'T ,T ( complex tori 1 C 1 and 2 C 2 with 1 v1 w1 and G 'T ,T ( w /v w /v 2 v2 w2 are conformally equivalent iff 1 1 and 2 2 are on the same orbit under the modular group SL(2, Z). To give a sketch proof, we notice ﬁrst that up to conformal equivalence, a 2 torus can be realized as C/'T1,Tτ(, where τ ∈ H . Assume now that f /'T ,T (→ /'T ,T ( we have a conformal equivalence : C 1 τ1 C 1 τ2 be- 2 tween two tori given by τ1 and τ2 in H . Since both tori are obtained from fundamental parallelograms in C by side-pairing transformations, it is clear that f can be “lifted up” to a conformal equivalence ˜ ˜ ˜ f : C → C satisfying f(0) 0 and the relation π2 ◦f f ◦π1, where π → /'T ,T ( π → /'T ,T ( 1 : C C 1 τ1 and 2 : C C 1 τ2 are natural projections. Complex calculus tells us that a conformal equivalence of C is linear, so that we have

f(z)˜ αz, z ∈ C, for some α ∈ C. (This follows since a conformal equivalence of C has a removable singularity at infinity, so that it extends to a conformal equivalence of the Riemann sphere.) The commutation relation for f and f˜ above implies that f˜ maps any linear combination of 1 and τ1 with integer coefficients to a linear combination of 1 and τ2 with integer coefficients. We thus have ˜ f(1) α a + bτ2, ˜ f(τ1) ατ1 c + dτ2,

6In topology, the deﬁnition of simply connectedness requires a short detour into homotopy theory. For- tunately, simply connectedness of a domain in the complex plane C is equivalent to connectedness of its complement in the extended plane Cˆ . 7See H. Farkas and I. Kra, Riemann Surfaces, Springer, 1980. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

188 15. Riemann Surfaces

for some a, b, c, d ∈ Z. Since f˜ is invertible, we have ad − bc ±1. Solving for τ1, we obtain

c + dτ2 τ1 , a + bτ2

so that τ1 is in the SL(2, Z)-orbit of τ2. (Note that ad − bc 1, since 2 both τ1 and τ2 are in H .) ♥ We turn now to the most important case of Fuchsian groups acting on the hyperbolic plane H2. Discreteness of the Fuchsian group G on H2 implies that H2/G is a Riemann surface with analytic projection π : H2 → H2/G. The proof of this is a souped up version of the one we just did for discrete translation groups for C. We omit the somewhat technical details. It is more important for us that a rich source of Riemann surfaces can be obtained this way. (In fact, all Riemann surfaces arise as quotients; this is “uniformization,” a more advanced topic.) Instead, we look at the examples of Fuchsian groups obtained in the previous section and ﬁnd the corresponding Riemann surface by looking at the side-pairing transformations on the fundamental hyperbolic polygon.

Example 1 G 'g( with g(z) z + 1, z ∈ H2. The modiﬁed exponential map z → exp(2πiz), z ∈ H2, is invariant under G, so that it projects down to H2/G and gives a conformal equivalence between H2/G and the punctured disk D2 −{0} (Figure 15.8). This example gives us the clue that parabolic elements in G are responsible for punctures in H2/G. (In general,

Figure 15.8 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

15. Riemann Surfaces 189 there is a one-to-one correspondence between the punctures of H2/G and the conjugacy classes of parabolic elements in G.)

Example 2 Let k>1 and G 'g( with g(z) kz, z ∈ H2. The Rie- mann surface H2/G is conformally equivalent to the annulus 2π2/ log k Ar {z ∈ C | 1 < |z|

˜ ˜ 2 f(k1z) k2f (z), z ∈ H ,

2 2 2π / log r1 2π / log r2 where k1 e and k2 e . Complex calculus (the Schwarz lemma) tells us that a conformal equivalence of H2 is necessarily Mobius,¨ so that az + b f(z)˜ ,ad− bc , a,b,c,d ∈ . cz + d 1 R

Combining this with the commutation relation above, we see that b 0, k1 k2, and hence r1 r2.

Example 3 Let n ∈ N and G 'g( with

cos(π/n) · z − sin(π/n) g(z) ,z∈ H2. sin(π/n) · z + cos(π/n)

This is best viewed on D2 where G is generated by the Euclidean rotation R2π/n around the origin (Figure 15.9). n The map z → z is invariant under 'R2π/n(, projects down to the quotient, and deﬁnes a conformal equivalence between H2/G and the unit disk D2. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

190 15. Riemann Surfaces

2π/n

Figure 15.9

Example 4

Let k>1 and G 'g1,g2( with g1(z) −1/z and g2(z) kz, z ∈ H2. A somewhat involved argument shows that H2/G is conformally equivalent to the unit disk.

Example 5 Let G SL(2, Z) be the modular group. The Riemann surface H2/G is conformally equivalent to C. This example is important for the Riemann moduli of tori. In fact, we now see that H2/SL(2, Z) and thereby C “parametrizes” the set of conformally inequivalent tori.

Example 6

Let G 'g1,g2(, where 3z + 4 g (z) and g (z) 2z, z ∈ H2. 1 2z + 3 2 Looking at the fundamental set, we see that H2/G is conformally equivalent to the punctured sphere; that is, to C.

Finally, if a Fuchsian group G has a fundamental polygon with 4p sides (and G identiﬁes the opposite sides), then H2/G is conformally equivalent to a compact genus p Riemann surface, or more plainly, a “torus with p holes.” As an example, use the four Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

15. Riemann Surfaces 191

Figure 15.10 hyperbolic side-pairing transformations for the hyperbolic octagon at the end of Section 14 as a fundamental domain of a Fuchsian group and realize that H2/G is a genus 2 Riemann surface (Figure 15.10). Toclose this section, we make a note on the origins of the theory of Riemann surfaces. We saw that the exponential map exp : C → C gives rise to a conformal equivalence between the cylinder and the punctured plane, with inverse being the complex logarithm

log w log |w|+i arg w, w 0.

If we reject the cylinder as range and keep log to be defined on the punctured plane C −{0}, then it is inevitably multiple valued. How can we get around this difficulty? Well, since log is multiple valued, the domain C−{0} has to be replaced by a Riemann surface on which it becomes single valued. The new domain has to have infinitely many layers of C−{0}, so that it must look like an infinite staircase denoted by St∞ (see Color Plate 2b). With this, we see that exp : C → St∞ is a conformal equivalence and that the Riemann surfaces are simply connected (no holes!). n The situation is similar√ for the power function z → z . In this case, the inverse w → n w is n-valued (FTA!), so that the finite staircase Stn will do. Color Plates 3a–b depict the cases n 2 and n 3. (Notice that St2 and St3 seem to have self-intersections. But remember, this is due to our limited 3-dimensional vision; as a n 4 matter of fact, Stn, the graph of z → z , lives in C × C R , and for a surface in 4 dimensions there is plenty of room to avoid self-intersections!) Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

192 15. Riemann Surfaces

♠ Given a monic complex polynomial P of degree n, by cutting 8 and pasting,√ a compact Riemann surface MP can be constructed on which P is single-valued and analytic. P can be assumed to have distinct roots, since the square root of a double root factor is single-valued. MP comes equipped with an analytic projection ˆ π : MP → C, and π is a twofold branched covering with branch points above the roots of P (and ∞ for n odd). If z ∈ C is away from the roots of P, then the two points π−1(z) correspond to ± P(z).We will construct MP explicitly for n ≤ 4. From this the general case will follow easily. If P is monic and√ linear with root a ∈ C, then the Riemann surface MP of P(z) z − a is essentially St2 (See Color Plate 3a) centered at a. MP is obtained by stacking up two copies of Cˆ , making in each copy a (say) radial cut from a to ∞, and 9 ˆ ﬁnally, pasting the four edges crosswise. The map π : MP → C corresponds to vertical projection, and this is a twofold branched√ covering with branch points above a and ∞. The function 1/ P has a simple pole above a and a simple zero above ∞. Notice that MP is ˆ ˆ conformally equivalent to C, and with the identiﬁcation MP C, π becomes the map z → (z − a)2, z ∈ Cˆ . We now realize that we met this a long time ago in the proof of the FTA! The situation for quadratic P with distinct roots a, b ∈ C is similar. The Riemann surface MP for P(z) (z − a)(z − b) is obtained by cutting the two copies of Cˆ by the line segment (actually, any smooth curve) connecting a and b, and pasting the four edges crosswise. The√ map π has branch points above a and ˆ b. The double-valued P lifted up along π : MP → C becomes single-valued on MP , since the winding number (cf. Problem 3 of Section 8) of a closed curve that avoids the cuts is the√ same with respect to the points above a and b. The function 1/ P has simple poles at the points above a and b and simple zeros at the two ˆ points above ∞. Once again, MP is conformally equivalent to C.We also see that the quadratic case can be reduced to the linear case by sending b to ∞ by a linear fractional transformation. Analyti-

8In the rest of this section, we describe Siegel’s approach to the Weierstrass ℘-function. For details, see C.L. Siegel, Topics in Complex Function Theory, Vol. I, Wiley-Interscience, New York, 1969. 9For pasting topological spaces in general, see “Topology” in Appendix C. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

15. Riemann Surfaces 193 z → /z + b (z − a)(z − b) cally,√ the substitution 1 transforms to (1/z) b − a z + 1/(b − a). It is rewarding to take a closer look at the particular case P(z) z2 − 1. Let the cuts be√ the line segments that connect ±1inthe ˆ two copies of C. Since P becomes single-valued on MP ,wecan consider the line integral

dz I(C) , C P(z) where the curve C emanates from a fixed point p0 ∈ MP and terminates at a variable point p ∈ MP. Although MP is simply con-√ nected, the integral I depends on C due to the simple poles of 1/ P above a and b. (To be precise, by the monodromy theorem, the line integral I(C) with respect to a curve C that avoids a, b, and ∞ depends only on the homotopy class of C in MP −{a, b, ∞}.) To make the integral I path-independent, we have to construct a Riemann surface M above MP by taking infinitely many copies of MP and then cutting and pasting, following the recipe that the integration prescribes (counting the residues a` la Cauchy). The prescription in question becomes more transparent when we realize that the√ inverse of√ the complex sine function is the antiderivative of 1/ 1 − z2 i/ z2 − 1, so that M should be the Riemann surface of sin−1! The sine function itself is given by the Euler formula for complex exponents: eiz − e−iz sin z . 2i This is one-to-one on any vertical strip (k − 1/2)π < !(z) < (k + 1/2)π, k ∈ Z, and each strip is mapped onto the whole complex plane with cuts (−∞, −1) and (1, ∞) along the real axis (cf. Problem 2). The line !(z) (k + 1/2)π corresponds to the two edges of the positive cut if k is even, and to the negative cut if k is odd. Thus, the Riemann surface on which the inverse of sine is single-valued is obtained from infinitely many copies of the complex plane with the cuts (−∞, −1) and (1, ∞) as above, and the even and odd layers connect each other alternately. If P is cubic with three distinct roots a, b, c ∈ C, then we group a, b, c, and ∞ into two pairs, connect them by disjoint smooth Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

194 15. Riemann Surfaces

curves, cut the two copies of Cˆ along these curves, and join the edges crosswise. This time MP becomes conformally equivalent to a complex torus. (This can be seen most easily by arranging the cuts in different hemispheres of S2 Cˆ , cutting one of the copies of S2 along the equator, pasting ﬁrst along the branch cuts crosswise, and ﬁnally pasting the cut equators back together.) As before, ˆ π : MP → C is an analytic twofold√ branched covering with branch points above√a, b, c, and ∞, and P is single-valued on MP . The function 1/ P has simple poles at a, b, c and a triple zero at ∞. By a linear change of the variables, we can put P in the classical Weierstrass form

3 P(z) 4z − g2z − g3. Since P has distinct roots, the discriminant δ of P (cf. Section 6) is nonzero: 1 δ (g3 − 27g2) 0. 16 2 3 The line integral

dz I(C) 3 C 4z − g2z − g3 is called an elliptic integral of the ﬁrst kind. As before, the monodromy theorem says that for a curve C that avoids a, b, c, the line integral I(C) depends only on the homotopy class of C in MP −{a, b, c}. In addition, being a complex torus, MP itself has nontrivial topology. Instead of cutting and pasting we unify these path-dependencies into a single concept. We collect the values of the line integral I(C) for all closed curves C (based at a ﬁxed point p0 away from a, b, c) and call them periods. By additivity of the integral, the set LP of all periods forms an additive subgroup of C.In fact, LP is a 2-dimensional lattice in C. (This is because the funda- 2 mental group π1(MP,p0) is Z .) We call LP the period lattice.Wesee that in order to make the line integral I(C) dependent on the terminal point of C only, we need to consider the value of I(C) modulo the period lattice LP. In other words, I effects a conformal equivalence from the Riemann surface MP to the complex torus C/LP. The inverse of this map is the Weierstrass ℘-function, which is best Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

15. Riemann Surfaces 195 viewed as an analytic map lifted from C/LP to C. By deﬁnition, ℘ satisﬁes the differential equation

2 3 (℘ ) 4℘ − g2℘ − g3,w∈ C. (What is the analogue for the sine function?) Moreover, ℘ is doubly periodic with periods in LP :

℘(w + ω) ℘(w), w ∈ C,ω∈ LP . This is all very elegant but does not give ℘ in an explicit form. To get to this, we ﬁrst notice that ℘ must have poles (Liouville’s theorem). In fact, ℘ is the simplest doubly periodic function. After a detailed analysis we arrive at the partial fractions expansion ℘(w) 1 + 1 − 1 . w2 (w − ω)2 ω2 ω∈LP ,ω 0

Notice that ℘ has double poles at the lattice points in LP , and the difference is needed to ensure convergence. (The series converges absolutely and uniformly on any compact subset in C−LP.) Finally, we note that the lattice LP itself determines the coefficients g2 and g3 in the classical Weierstrass form above. In fact, we have 1 1 g 60 and g 140 2 ω4 3 ω6 ω∈LP ,ω 0 ω∈LP ,ω 0 (cf. Problem 3). We now take a closer look at the differential equation that ℘ satisfies. This looks very familiar! In fact, we immediately notice that, for any w ∈ C, the pair (℘(w), ℘(w)) satisfies the equation

2 2 y P(x) 4x − g2x − g3.

2 This is our old friend the elliptic curve Cf , f(x, y) y −P(x), which we studied in Section 3! The only difference is that here x and y are complex variables, so that our elliptic curve sits in C2 (rather than in R2). In fact, we also need to recall the points at inﬁnity that needed to be attached to R2. In our case, an ideal point is a pencil of complex lines in C2, and, in analogy with the real case, C2 with the ideal points becomes the complex projective plane CP2. Thus the 2 equation above deﬁnes a complex elliptic curve Cf (C) in CP , and Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

196 15. Riemann Surfaces

we have an analytic map 2 (℘, ℘ ) : C/LP → Cf (C) ⊂ CP . It is not hard to show that this map is a conformal equivalence. With this, we obtain that Cf (C) is a complex torus. We now realize that Figures 3.10 to 3.17 depict various slices of this torus; the only visual problem is the absence of the vertical inﬁnity! We want to assert that our conformal equivalence is actually an algebraic isomorphism. We have a little technical trouble here. It is clear how to add points in C/LP , but we deﬁned addition on elliptic curves only for real coordinates and not, in general, for points on Cf (C). This problem can be easily resolved. All we need to do is to work out the algebraic formulas for the geometric rule for the addition (the chord method), and the formulas will automatically extend to the complex case (cf. Problem 17 of Section 3). Notice also that the algebraic isomorphism between C/LP and Cf (C) is nothing but the classical addition formula for the ℘-function: 2 1 ℘ (w1) − ℘ (w2) ℘(w1 + w2) −℘(w1) − ℘(w2) + . 4 ℘(w1) − ℘(w2)

Remark. In Section 3, we defined addition of points on an elliptic curve using geometry (chord method). In an analytical approach10 we could first define the Weierstrass ℘-function, then use ℘ to establish the conformal equivalence of an elliptic curve (over C) with a complex torus (by choosing the lattice suitably), and finally define addition on the elliptic curve by carrying over the obvious addition on the torus to the curve, or equivalently, declaring the conformal equivalence to be an algebraic isomorphism. As expected, the case of a quartic polynomial P with distinct roots a, b, c, d ∈ C can be reduced to the cubic case. The complex algebraic curve defined by y2 P(x) is birationally equivalent to an elliptic curve. Without going into details, we mention yet another connection. The elliptic integral for a quartic P can be put into the Legendre

10 This is followed in Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer, 1993. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 197 form

dz I(C) , C (1 − z2)(1 − kz2) and this gives the Schwarz–Christoffel formula for a conformal map of the rectangle with vertices (±1, ±k) onto the upper half-plane H2! The construction of the Riemann surface MP for P a monic polynomial of any degree n (with distinct roots) can be easily generalized from the particular cases above. As before, we group the roots in pairs with ∞ added if n is odd, connect the pairs of points with disjoint smooth curves, make the [(n + 1)/2] cuts on each copy of Cˆ along the curves, and, ﬁnally, join the corresponding cuts crosswise. The Riemann surface MP is conformally equivalent to a torus with [(n − 1)/2] holes.

Problems

z n 1. Derive Euler’s formula for complex exponents using e limn→∞(1 + z/n) , where z is a complex number. 2. Prove the basic identities for the complex sine function and derive its mapping properties stated in the text. 3. Use (the derivative of) the geometric series formula for 1/(w − ω)2 − 1/ω2 to derive the expansion 1 ℘(w) + 3G w2 + 5G w4 + 7G w6 +···, w2 4 6 8 where G 1 . k ωk ω∈LP ,ω 0

Substitute this into the differential equation of ℘ to obtain the formulas for g2 and g3 stated in the text. 4. Work out the side pairing transformations for the hyperbolic octagon in the last example of Section 14 and verify that pasting11 gives the two-holed torus.

11 For a different cutting-and-pasting construction of hyperbolic octagons and dodecagons, cf. D. Hilbert and S. Cohn-Vossen, Geometry and Imagination, Chelsea, New York, 1952. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

198 15. Riemann Surfaces Web Site

1. www.geom.umn.edu/∼banchoff/script/CFGPow.html Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... General Surfaces . . . 16 . SECTION .

♥ All Riemann surfaces can be listed as either S2, C/G with G a translation group acting on C or H2/G with G a Fuchsian group acting on H2. The meager possibilities for S2 and C tempt us to think that we may obtain surfaces more general than Riemann surfaces by relaxing some of the conditions on the acting discrete group G. What should we expect to give up to arrive at these more general surfaces? To answer this question, we need to reconsider the definition of a Riemann surface. The main restriction there came from re- −1 quiring the chart-changing transformation ϕk ◦ ϕj , j, k ∈ N,tobe differentiable in the complex sense since, as we have seen, there are many transformations that are nice (in the real sense) but fail to be analytic. For example, any Riemann surface carries an orientation given a Riemann surface, we know how to rotate positively around a point. This is because the local orientations given by the charts ϕj patch up to a global orientation, since changing charts amounts −1 to performing ϕk ◦ ϕj , and this, being conformal, preserves the local orientations. ♠ It is now clear how to define the concept of a general surface. Just repeat the definition of a Riemann surface (replacing C by R2), −1 and instead of saying that each ϕk ◦ ϕj , j, k ∈ N, is analytic, we just 199 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

200 16. General Surfaces

require this to be differentiable1 in the real sense! We now expect to obtain interesting new surfaces of the form R2/G, where G is a discrete group of diffeomorphisms acting on R2. The classification of these quotients is possible but still a formidable task. Since we wish to stay in geometry, we require the elements of G to be isometries. In addition, we will assume that each element of G (that is not the identity) acts on R2 without fixed points. If you worry about this condition being too restrictive, recall z → zn, z ∈ C, from the proof of the FTA,where the corresponding quotient did not give anything new but S2! Your worry, however, is not unfounded. Excluding transformations with fixed points makes the projection map π : R2 → R2/G a simple covering, while at the fixed points of the elements in G, π would “branch over”—a much more interesting phenomenon. Our reason for leaving out the branched coverings is mostly practical; otherwise we would never get out of two dimensions and on to later parts of the Glimpses! We now discuss examples. First assume that G is one of the seven frieze groups (see Section 10). Since G can only contain translations and glides, this leaves us only the first and seventh types. The first frieze group is generated by a translation, and the resulting surface R2/G is a cylinder, a Riemann surface discussed in Section 15. Assume now that G is of the seventh type; that is, it is generated by a single glide Gl,v. We may assume that l is the first axis and v is 2π times the first unit vector: v (2π, 0) ∈ R2. A fundamental domain F0 for G is the vertical strip (0, 2π)×R, and the side-pairing transformation

Gl,v|{0}×R : {0}×R →{2π}×R is given by

Gl,v(0,y) (2π, −y), y ∈ R. The quotient R2/G is called the infinite Möbius band. It can be visualized in the following way: Consider the unit circle S1 ⊂ R2 ⊂ R3 (in the plane spanned by the first two axes) and the line parallel to the third axis in R3 through (1, 0) ∈ S1. If you slide the line along S1 by keeping it perpendicular to R2 all the time, it sweeps

1See “Smooth Maps” in Appendix D. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

16. General Surfaces 201 an ordinary cylinder. Now slide the line along S1 with unit speed and, while sliding, rotate it by half of that speed. (Do not worry about self-intersections at this point; in fact, if you rotate the line in an extra 2-dimensional plane perpendicular to S1 ⊂ R2, then there will be no self-intersections, and the Mobius¨ band will be imbedded in R2 × R2 R4!) Upon going around S1 once, we complete a half- turn of the line. Now, what the line sweeps is the Mobius¨ band. It is much easier to visualize this when we consider the action of the glide only on a finite strip R × (−h/2/ 2) of height h>0. We obtain the finite Möbius band. A note about orientability: We observed above that for a Riemann surface, changing the charts from ϕj : Uj → C to ϕk : Uk → C −1 amounts to performing ϕk ◦ ϕj , and this, being complex differentiable, is always orientation preserving. In the case of general surfaces, we see that the surface is orientable if there exists an atlas −1 in which every chart-changing transformation ϕk ◦ ϕj is orientation preserving2. It is easy to see that such an atlas cannot exist on aMobius¨ band, and therefore it is not orientable. The same applies to all surfaces that contain the Mobius¨ band, and this observation is sufficient for all the examples that follow. A note before we go any further: The square of a glide is a translation, and the quotient of R2 by a group generated by a single translation is the cylinder. Thus, the map 2 2 2 R /'Gl,v(→R /'Gl,v( 2 that associates to the orbit 'Gl,v((p) the orbit 'Gl,s((p) is two-to-one (that is, every point on the range has exactly two inverse images). We obtain that the cylinder is a twofold cover of the Mobius¨ band! ♥ We now turn to crystallographic groups acting on R2. Toobtain a surface that has not been listed so far, we assume that G contains a glide. A quick look at the seventeen crystallographic groups shows that we are left with only the case when G is generated by two parallel glide rections. (Note that the case of two perpendicular G 'G ,G ( v ·v glides cannot occur.In fact, if l1,v1 l2,v2 with 1 2 0, then (v −v )/ G ◦G T ◦R ◦ 2 1 2 is a fixed point of the composition l2,v2 l1,v1 v2 l2 R ◦ T R ◦ R H G 'G ,G ( l1 v1 , since l2 l1 is a half-turn.) We set l1,v1 l2,v2 ,

2See “Smooth Maps” in Appendix D. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

202 16. General Surfaces

Figure 16.1

where l1 R ×{0},l2 R ×{h}, h>0, and v1 v2 (2π, 0). A fundamental domain is given by F0 (0, 2π) × (−h, h) (Figure 16.1). The side-pairing transformations are illustrated in Figure 16.2. G The first side-pairing transformation is the glide l1,v1 , and the −1 second is the translation G ◦ Gl ,v . The resulting quotient C/G l2,v2 1 1 is called the Klein bottle, denoted by K2. We claim that K2 is obtained by pasting two copies of the Mobius¨ band together along their boundary circle (of perimeter 4π). Indeed, cut the fundamental domain F horizontally along the lines R ×{h/2} and R ×{−h/2} (Figure 16.3). This cut is a topological circle on K2 since (0/ 2) is identified with (2π, −h/2) and (0, −h/2) is identified with (2π, h/2). The middle portion gives a Mobius¨ band. The upper and lower portions −1 (identified by G ◦ Gl ,v ) first give a rectangle (Figure 16.4), the l2,v2 1 1 two vertical sides are identified again by a glide (from appropri- G ate restrictions of l1,v1 ), and this gives another Mobius¨ band! A somewhat more visual picture of K2 is shown in Color Plate 4a. Here, instead of rotating a straight segment, we rotated two halves of the lemniscate (Figure 16.5) to obtain two topological copies of the Mobius¨ band. Then pasting is no problem! No- tice that since K2 contains a Mobius¨ band (actually, it contains

Figure 16.2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

16. General Surfaces 203

Figure 16.3

Figure 16.4 two), it is nonorientable. Notice also that K2 can be covered by a torus with a twofold covering. In fact, the torus in question is 2 −1 2 R /'G ◦ Gl ,v ,G (, and the fundamental domain of the torus l2,v2 1 1 l1,v1 cover is obtained by “doubling” F in the horizontal direction. ♦ Finally, we consider discrete groups acting on the sphere S2. Our condition that the isometries act on S2 without ﬁxed points imposes a severe restriction. The following theorem is essentially due to Euler:

Figure 16.5 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

204 16. General Surfaces

Theorem 10. Let S:S2 → S2 be a nontrivial isometry. If S has a fixed point on S2, then S is the restriction of a spatial reflection or a spatial rotation. If S 2 2 has no fixed point on S , then there exists p0 ∈ S such that S(p0) −p0, S leaves the great circle C orthogonal to p0 invariant, and S acts on C as a rotation.

Proof. First note that S is the restriction of an orthogonal transformation U : R3 → R3. This follows from the following argument: 2 Let p1,p2,p3 ∈ S be points not on the same great circle. Then there exists an orthogonal transformation U : R3 → R3 such that U(pl) S(pl), l 1, 2, 3. (This is because the spherical triangles Kp1p2p3 and KS(p1)S(p2)S(p3) are congruent.) The composition −1 2 U ◦ S is an isometry on S and ﬁxes pl,l 1, 2, 3. As in the Euclidean case, it follows that U −1 ◦ S is the identity, so that S U on S2. U is represented by a 3 × 3 orthogonal matrix. To look for eigenvectors p and eigenvalues λ for U, we solve the equation U(p) λ · p. We know from linear algebra that λ satisﬁes the characteristic equation det(U − λI) 0. Since U isa3× 3 matrix, this is a cubic polynomial in λ. Every cubic (in fact, odd degree) polynomial P(λ) has a real root λ0 (see Problem 4 of Section 8). Let 3 p0 ∈ R be an eigenvector of U corresponding to the eigenvalue λ0. Since U preserves lengths, we have λ0 ±1, so that U(p0) ±p0. 2 Let C ⊂ S be the great circle perpendicular to p0. C is the intersec- p⊥ p S2 tion of the plane 0 perpendicular to 0 and . We now claim that U p⊥ w leaves 0 invariant. This follows from orthogonality. In fact, if is perpendicular to p0, then U(w) is perpendicular to U(p0) ±p0, U p⊥ and the claim follows. restricted to 0 is a linear plane isometry, so it must be a rotation or a rection. By looking at the possible combinations, we see that the theorem follows.

Let G be a discrete group of isometries of S2 and assume that each nonidentity element of G has no ﬁxed point. We consider only the simplest case in which G is cyclic and generated by a single element g ∈ G. Discreteness, along with Theorem 10, implies that g is a spatial rotation with angle 2π/n, n 2, followed Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

16. General Surfaces 205 by spatial rection in the plane of the rotation (perpendicular to the axis of the rotation). It is now a simple fact that S2/G is topologically the same for all n. (Look at a fundamental domain bounded by two meridians of longitude!) We set n 2. Then g becomes the antipodal map −I : S2 → S2, −I(p) −p, and G {±I}. The quotient S2/{±I} is called the real projective plane denoted by RP2. In the standard model for the real projective plane, projective points are interpreted as lines through the origin in R3, and projective lines as planes containing the origin of R3. Since the origin is a multiple intersection point, it is deleted from the model. Since every two projective lines intersect, we obtain a model for elliptic geometry. Algebraically, we let ∼ be the equivalence relation on 3 R −{0} deﬁned by p1 ∼ p2 iff p2 tp1 for some nonzero real t. The equivalence class containing p ∈ R3 −{0} is the projective point that corresponds to the line R · p passing through p and with the origin deleted. If p (a,b,c), then this projective point is classically denoted by [a : b : c]. We also say that a, b, c are the homogeneous coordinates of the projective point with the understanding that for t nonzero, ta, tb, tc are also projective coordinates of the same projective point. By deﬁnition, the set R3 −{0}/ ∼ of equivalence classes is the real projective plane RP2. Associating to a nonzero point the equivalence class it is contained in gives the natural projection R3 −{0}→RP2. Since we have no space-time here to explore the sublime beauty of projective geometry, we will understand RP2 in topological terms only. The idea is to replace R3 −{0} by the unit sphere S2 and to consider the intersections of projective points and lines with S2. Each projective point intersects S2 at an antipodal pair of points. Moreover, knowing this pair, one can reconstruct the projective point by considering the Euclidean line through them. The intersection of a projective line with S2 is a great circle (which we see as a better representative of a line than a plane anyway). Every projective line is thus a topological circle.3

3I heard the following story from a reliable source: A desperate student asked a professor what he could do for a passing grade in geometry. “Draw a projective line,” was the answer. The student took the chalk and started drawing a horizontal line. “Go on,” said the professor when he got to the end of the chalkboard. So Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

206 16. General Surfaces

Since every projective point corresponds to a pair of antipodal points of S2, a topological model of the projective plane is obtained by identifying the antipodal points with each other: RP2 S2/{±I}. We also see that the identification projection π : S2 → RP2 is a twofold cover. How can we visualize this? Think of S2 as being the Earth and divide it into three parts with the Arctic and Antarctic Circles. Between these parallels of latitude lies a spherical belt that we further divide by the 0◦ and 180◦ meridians of longitude. Since the two spherical caps and the two halves of the belt are identified under −I, we keep only one of each (see Color Plate 4b). The longitudinal sides of the half-belt are identified under −I, and we obtain a Mobius¨ band. The cap is attached to this. Thus, RP2 is a Mobius¨ band and a disk pasted together along their boundaries! In particular, RP2 is nonorientable. The real projective plane has another classical model, based on an extension of the Euclidean plane by adding a set of so-called ideal points. To describe these, consider the equivalence relation of parallelism on the set of all straight lines in R2. We call an equivalence class (that is, a pencil of parallel lines) an ideal point. We define RP2 as the union of R2 and the set of all ideal points. Thus, a projective point is either an ordinary point in R2 or an ideal point given by a pencil of parallel lines. A projective line is the union of points on a line l plus the ideal point given by the pencil of lines parallel to l. There is a single ideal line, filled by all ideal points. Incidence is defined by (set-theoretical) inclusion. How does the algebraic description fit in with the geometric description of RP2 as the extension of R2 with ideal points corresponding to pencils of parallel lines? If c 0, then [a : b : c] and [a/c : b/c : 1] denote the same projective point. Hence, adjusting the notation (or setting c 1), the Euclidean point (a, b) ∈ R2 can be made to correspond to the projective point [a : b :1]∈ RP2, and this correspondence is one-to-one with the plane of ordinary points in RP2. Thus, the ordinary points in RP2 are exactly those that have nonzero third homogeneous coordinates. For an ideal point [a : b :0]∈ RP2,we he continued drawing on the wall, went out of the classroom, down the hallway and out to the street. By the time he got back from his roundabout tour, the professor had already marked in the passing grade. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

16. General Surfaces 207 either have a 0 and [a : b :0] [1 : m : 0], m b/a, so that this may be thought to represent the pencil of parallel lines with common slope m,ora 0 and [0 : b :0] [0:1:0],which represents the pencil of vertical lines. It is clear that two distinct projective lines always intersect. (In axiomatic treatment, this is called “the elliptic axiom.”) Indeed, they either meet at an ordinary point or (their Euclidean restic- tions) are parallel, in which case they meet at the common ideal point given by these lines. This model is the same as the topological model given by RP2 S2/{±I}. Indeed, cut a unit sphere into two hemispheres along the equator and keep only the southern hemisphere H. Let H sit on R2 with the south pole S touching the plane (see Figure 16.6). Apply stereographic projection from the center O of H. The points in R2 correspond to points of the interior of H (i.e., H without the boundary equatorial circle). Each line in R2 corresponds to a great semicircle on H ending at two antipodal points of the equatorial circle. These endpoints, identiﬁed by the antipodal map, give the single ideal point of the projective extension of the line. Thus, the ideal points correspond to our horizontal view, and the boundary equatorial circle (on H modulo the antipodal map) gives the ideal projective line. You are now invited to check the basic properties of this model.

Figure 16.6 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

208 16. General Surfaces Problems

1. Fill in the details in the following argument, which gives another proof of Theorem 10. (a) Show that if S leaves a great circle C on S2 invariant, then each of the 2 two p0 ∈ S perpendicular to C satisﬁes S(p0) ±p0. (b) To construct C, consider the function f : S2 → R deﬁned by f(p) the spherical distance between p and S(p), p ∈ S2. Assume that 0

cos(θ) − sin(θ) sin(θ) cos(θ)

corresponding to planar rotation with angle θ (and, for n odd, by a single 1 × 1 block with unit entry). 3. Define and study Iso (RP2). 4. Make a topological model of the Klein bottle K2 from a finite cylinder bent into a half-torus by pasting the two boundary circles together appropriately. 5. Define the real projective n-space RPn using homogeneous coordinates. (a) Show that RP1 is homeomorphic to S1. (b) Imbed RPn−1 into RPn using the inclusion Rn ⊂ Rn+1 given by augmenting n-vectors with an extra zero coordinate. What is the difference RPn − RPn−1?

Web Site

1. vision.stanford.edu/∼birch/projective/node3.html Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... The Five Platonic . . . . . Solids . 17 . SECTION .

♣ A triangle in the plane can be thought of as the intersection of three half-planes whose boundary lines are extensions of the sides of the triangle. More generally, a convex polygon is deﬁned as a bounded region in the plane that is the intersection of ﬁnitely many half-planes. A convex1 polygon has the property that for each pair p1,p2 of points of the polygon, the segment connecting p1 and p2 is entirely contained in the polygon (Figure 17.1). The regular n-sided polygon is a primary example of a convex polygon. It is distinguished among all convex polygons by the fact that all its sides and angles are congruent.

Figure 17.1

1Nonconvex objects surround us. Next time you eat breakfast, take a closer look at your croissant or bagel. 209 This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

210 17. The Five Platonic Solids

In a similar vein, in space, a convex polyhedron K is defined as a bounded region in R3 that is the intersection of finitely many half-spaces. The notion of convexity carries over to three (and in fact any) dimensions. The part of the boundary plane of the half- space participating in the intersection that is common with the polyhedron is called a face of K. Any common side of two faces is an edge. The endpoints of the edges are the vertices of K. Regularity of convex polyhedra, although simple in appearance, is not so easy to define. It is clear that the faces of a regular polyhedron must be made up of regular polygons all congruent to each other. That this is not enough for regularity is clear when one considers a double pentagonal pyramid called Bimbo’s lozenge (Figure 17.2). At two vertices five equilateral triangles meet, while only four meet at the remaining five vertices. To define regularity, we turn to group theory. First we define Iso (R3), the group of isometries of R3 (in much the same way as we did for R2); a transformation S : R3 → R3 belongs to Iso (R3) if S preserves spatial distances: d(S(p), S(q)) d(p, q), p, q ∈ R3, where d : R3 × R3 → R is the Euclidean distance function: d(p, q) |p − q|,p,q∈ R3. As in two dimensions, we can easily classify all spatial isometries. Those that have a fixed point have been described in Theorem 10 (see Section 16). In fact, this result can be rephrased by saying that any element of Iso (R3) that leaves a point fixed is either a spatial rection or a rotatory reflection, a rotation followed by re-

Figure 17.2 This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 211

flection in the plane perpendicular to the rotation axis. (What is a rotatory half-turn?) Notice that Theorem 10 is actually stated in a bit stronger setting; for the conclusions all we need is the restriction 2 of the spatial isometry S to the unit sphere S (p0) with center p0,a fixed point of S. As in the plane, a spatial isometry S with no fixed points is the composition of a spatial translation Tv and another spatial isometry U that leaves the origin fixed; S Tv ◦ U, where v S(0).(U is linear, but we do not need this additional fact here.) Since we are in space, the possible outcomes of the composition of Tv and U depend on how the translation vector v relates to the rection plane or the rotation axis for U; think of the motion of the frisbee or uncorking a bottle! If U is a spatial rection, then S is a glide rection, a rection in a plane followed by a translation with vector parallel to the rection plane. (Indeed, this follows by v v + v T T ◦ T v v writing 1 2 and v v1 v2 , where 1 is parallel and 2 perpendicular to the rection plane. Notice the presence of the two-dimensional statement that any plane rection in a line followed by a translation with translation vector perpendicular to the rection axis is another rection; see Section 9.) If U is a rotation, then S Tv ◦ U is a screw displacement, a rotation followed by a translation along the rotation axis. (Once again this follows by decomposing v v1 + v2 as above, and using the two-dimensional statement, any plane rotation followed by a translation is another rotation; see Section 9.) Finally, piecing these arguments together to a single proof we obtain that if U is a rotatory rection, then so is S. (A spatial rection and a translation commute if the transla- T tion vector is parallel to the rection plane. Now distribute v1 and T U v2 to the rotation and rection part of .) Summarizing, every spatial isometry is either a rotation, a rection, a rotatory rection, a translation, a glide rection, or a screw displacement. As a byproduct we obtain that any spatial isometry is the composition of at most four spatial rections. (For example, a rotatory half-turn, the negative of the identity, is the composition of three rections in mutually perpendicular planes!) We want to consider a convex polyhedron K regular if we can carry each vertex of K to another vertex by a suitable spatial isometry in Symm (K), the group of spatial isometries that leave K This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

212 17. The Five Platonic Solids

invariant. In two dimensions, regularity of a polygon was certainly equivalent to this; think of the cyclic group of rotations leaving the regular n-sided polygon invariant (see Section 9). Similar symmetry conditions should be required for the edges and faces of K. Between vertices, edges, and faces there are incidence relations that essentially define K. To incorporate all these into a single symmetry condition, we introduce the concept of a flag in K as a triple (p,e,f)where p is a vertex, e is an edge, f is a face of K, and p ∈ e ⊂ f . We now say that K is regular if given any two flags (p1,e1,f1) and (p2,e2,f2) of K, there exists a spatial isometry S ∈ Symm (K) that carries (p1,e1,f1) to (p2,e2,f2); that is, S(p1) p2, S(e1) e2, and S(f1) f2. Let K be a convex polyhedron with vertices p1,...,pn. Define the centroid of K as

p +···+p c 1 n . n

We claim that every spatial isometry in Symm (K) ﬁxes c.Todo 2 3 this, we consider the function f : R → R deﬁned by f(p) n d(p, p )2 p ∈ 3 S ∈ (K) S j 1 j , R . Let Symm . Since permutes the vertices of K, {S(p1),...,S(pn)} {p1,...,pn}. It follows that f(S(p)) f(p), p ∈ R3. Indeed, we compute

n n 2 −1 2 f(S(p)) d(S(p), pj) d(p, S (pj)) j 1 j 1 n 2 d(p, pj) f(p). j 1

To show that S(c) c, we now notice that f has a global minimum

2Beautiful geometric arguments exist to prove this claim; see H.S.M. Coxeter, An Introduction to Geometry, Wiley, 1969. For a change, we give here an analytic proof. This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 213 at c. This follows by completing the square:3 n n 2 2 f(p) d(p, pj) |p − pj| j 1 j 1 n 2 2 (|p| − 2p · pj +|pj| ) j 1 n 2 2 n|p| − 2np · c + |pj| j 1 n 2 2 2 n|p − c| − n|c| + |pj| , j 1 where we used the definition of c and the dot product. Since the last two terms do not depend on p, it is clear that f(p) attains its global minimum where |p − c|2 does—that is, at p c. The composition f ◦ S−1 takes its global minimum at S(c). But f ◦ S−1 f , so these minima must coincide. S(c) c follows. If K is regular, its centroid has the same distance from each vertex. This is because regularity ensures that every pair of vertices can be carried into one another by a spatial isometry in Symm (K) that fixes c. The same is true for the edges and the faces of K.In particular, a sphere that contains all the vertices can be circumscribed around K. Notice that by projecting K from the centroid to the circumscribed sphere, we obtain spherical tessellations (see Problem 10). How many regular polyhedra are there? Going back to the plane is misleading; there we have infinitely many regular polygons, one for each positive integer n ≥ 3, where n is the number of edges or vertices. As we will show in a moment, there are only five regular polyhedra in R3. They are called the five Platonic solids (Fig- ure 17.3), since Plato gave them a prominent place in his theory of ideas.

3Did you notice how useful completing the square was? We used this to derive the quadratic formula, to ﬁnd the center of a hyperbolic circle, to integrate rational functions, etc. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

214 17. The Five Platonic Solids

Figure 17.3

Remark. The two halves of Bimbo’s lozenge are hiding in the icosahedron with a “belt” of ten equilateral triangles separating them! The belt conﬁguration (closed up with two regular pentagons) was called by Kepler a pentagonal antiprism. In a similar vein, an octahedron can be thought of as a triangular antiprism. (The pentagonal antiprism appeared about 100 years earlier as octaedron elevatum in Fra Luca Pacioli’s Da Divina Proportione, printed in 1509. This classic is famous for its elaborate drawings of models made by Leonardo da Vinci.) Following the “icosahedral recipe,” we can also insert a square antiprism between two square pyramids (two halves of an octahedron) and obtain a nonregular polyhedron with sixteen equilateral triangular faces (see Figure 17.4).

Figure 17.4 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 215

2π/a π/2 − π/a

Figure 17.5

We describe a regular polyhedron by the so-called Schlaﬂi¨ symbol {a, b}, where a is the number of sides of a face and b is the number of faces meeting at a vertex. Each face is a regular a-sided polygon. The angle between two sides meeting at a vertex is therefore π − 2π/a (Figure 17.5). (Split the polygon into a isosceles triangles by connecting the vertices of the polygon to the centroid.) At a vertex of the polyhedron, b faces meet. By convexity, the sum of angles just computed for the b faces must be < 2π.We obtain

b(π − 2π/a) < 2π.

Dividing by π, we obtain

b(1 − 2/a) < 2, or equivalently,

(a − 2)(b − 2)<4.

On the other hand, a, b > 2 by deﬁnition. A case-by-case check gives all possibilities for the Schli¨ symbol:

{3, 3}, {3, 4}, {4, 3}, {3, 5}, {5, 3}. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

216 17. The Five Platonic Solids

We could now go on and describe these solutions geometrically. Instead, for the moment, we take a brief look at the examples4 (but not the last column!) in Figure 17.6. Following Euler, we reﬁne the somewhat crude argument above to obtain the number of possible faces, edges, and vertices of each Platonic solid. We start with an arbitrary (not necessarily regular) convex polyhedron K.

Euler’s Theorem for Convex Polyhedra.5 Let V number of vertices of K; E number of edges of K; F number of faces of K. Then we have V − E + F 2.

Proof. We first associate to a convex polyhedron K a planar graph called the Schlegel diagram of K. To do this we use stereographic projection. Adjust K in space so that the top face is horizontal (that is, parallel to the coordinate plane spanned by the first two axes). Sit in the middle of the top face and look down to the transparent polyhedron K. The perspective image of the edges (the wireframe) gives a graph on the horizontal coordinate plane and defines a planar graph (with nonintersecting edges) if you are not too tall. Each face of K will correspond to a polygonal region in the plane bounded by the edges of the graph. We let the face you are sitting on correspond to the unbounded region that surrounds the graph. We can now analyze the last column of Schlegel diagrams of the five Platonic solids in Figure 17.6. Under stereographic projection, edges and vertices of K correspond to edges and vertices of the Schlegel diagram, so we begin to

4See H.S.M. Coxeter, Introduction to Geometry, Copyright 1969 by John Wiley & Sons, Inc. Reprinted by permission of John Wiley & Sons, Inc. 5This was known to Descartes, and according to widespread belief, to Archimedes as well. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 217

Figure 17.6 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

218 17. The Five Platonic Solids

suspect that Euler’s theorem must be valid in general for connected planar graphs, where any graph is defined by a finite number of points of the plane (called vertices) and a finite number of nonintersecting segments (called edges) connecting the vertices. The graph is further assumed to be connected, and it decomposes the plane into nonoverlapping regions. We now claim that Euler’s theorem is true for all connected planar graphs consisting of at least one vertex. To show this, we build a planar graph step by step, starting from a single vertex graph. For this trivial case the alternating sum V − E + F is 2, since V 1, E 0, and F 1. To build the graph, at each step we apply one of the following operations:

1. A new edge is added that joins an old vertex and a new vertex; 2. A new edge is added that joins two old vertices.

In each case, we have the following changes:

1. V + 1 → V, E + 1 → E, F → F; 2. V → V, E + 1 → E, F + 1 → F.

The alternating sum remains unchanged; Euler’s theorem follows!

We now return to our regular polyhedra with Schli¨ symbol {a, b} and use the additional information we just gained:

V − E + F 2.

The numbers in {a, b} relate to V, E and F by

2E aF.

Indeed, if we count the b edges at each vertex, we counted each edge twice. Similarly, if we count the a sides of each face, we again counted each edge twice. Combining these, we easily arrive at the following: a ab b V 4 ,E 2 ,F 4 . 2a + 2b − ab 2a + 2b − ab 2a + 2b − ab Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 219

We now look at Figure 17.6 again and see that the numbers of vertices, edges, and faces determined by these give6 the tetrahedron, cube, octahedron, dodecahedron, and icosahedron—the five Platonic solids! In ancient times, the existence of only five regular polyhedra called for much mysticism. In Plato’s Timaeus, the four basic elements—air, earth, fire, and water—were mysteriously connected to the octahedron, cube, tetrahedron, and icosahedron (in this order). To the dodecahedron was associated the entire Uni- verse. The latter is probably due to Timaeus of Locri, one of the earliest Pythagoreans. The twelve faces of the dodecahedron were believed to correspond to the twelve signs of the Zodiac. Figure 17.7 is adapted from a drawing by Kepler. Unlike the tetrahedron, the cube and the octahedron are common, basic structures for many crystals (such as sodium sulfantimoniate, common salt, and chrome alum). The occurrence of dodecahedral and icosahedral structures are rare in nonliving nature. However, these do occur in living creatures; for example, they are found in the skeletons of some microscopic sea animals called radiolaria.7 Moreover, a number of viruses such as the ade- novirus (which causes the flu and a host of other illnesses) have icosahedral structure. Let us now go back to mathematics and take a closer look at the Platonic solids. We have demonstrated above that there are only five regular polyhedra in space, but their actual existence was largely taken for granted. Of course, nobody doubts the existence of the regular tetrahedron, much less the cube, but how the faces of the dodecahedron, the icosahedron, and, to a lesser extent, the octahedron piece together remains to be seen. To reduce the number of cases, we now make some preparations and introduce the concept of reciprocal for regular polyhedra. Let {a, b}, a, b ≥ 3, be the Schli¨ symbol of a Platonic solid P whose existence we now assume. Let p be a vertex of P.Atp,

6It is perhaps appropriate to recall some of the Greek number preﬁxes here: 2 = di, 3 = tri, 4 = tetra, 5 = penta,6=hexa,7=hepta, 8 = octa, 9 = ennia, 10 = deca, 12 = dodeca, 20 = icosa. The cube does not ﬁt in; how would you rename it? 7See H. Weyl, Symmetry, Princeton University Press, 1952. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

220 17. The Five Platonic Solids

Figure 17.7. M. Berger, Geometry II, 1980, 32. Reprinted by permission of Springer-Verlag New York, Inc.

exactly b edges and faces meet. We denote the edges by e1,...,eb, and the faces by f1,...,fb in such a way that e1,e2 ⊂ f1, e2,e3 ⊂ f2,...,eb,e1 ⊂ fb (Figure 17.8). Let S ∈ Symm (P) be an isometry that carries the flag (p, e1,f1) into (p, e2,f2). (This is the first time when regularity comes in with full force!) Like every element of Symm (P), S fixes the centroid c of P, and thus it must fix the line l through c and p. (Notice that c and p must be distinct. Why?) S(e2) e3 since S(e2) is an edge Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 221

ee21

f 1 f f e3 2 b eb

Figure 17.8

2 of f2 S(f1) other than e2. Thus S (e1) S(e2) e3. Iterating, we j b obtain S (e1) ej+1, j 1,...,b− 1, and S (e1) e1. Thus S is a rotation of order b and axis l. It is now easy to see that the cyclic group 'S( generated by S is precisely the subgroup of Symm (P) of direct isometries that leaves p fixed. Armed with this description of Symm (P) at a vertex, we are now ready to define the reciprocal of P. Let P have Schli¨ symbol {a, b}. Consider a vertex p of P and denote by e1,...,eb the edges of P that meet at p. Let m1,...,mb be the midpoints of e1,...,eb. Since the endpoints of e1,...,eb other than p are in the orbit of 'S(,soarethe midpoints. Thus, m1,...,mb are the vertices of a regular b-sided polygon, and this polygon is in a plane perpendicular to the axis of rotation l of S (Figure 17.9). The polygon with vertices m1,...,mb is called the vertex figure of p at P. There is a vertex figure for each vertex of P, of which we have V in number. The V planes of the vertex figures enclose a polyhedron P0 that is called the reciprocal of P. What is the Schli¨ symbol of P0? Looking at the local picture of two adjacent vertices of P, we see that the edges of P0 bisect the edges of P at right angles. Of these bisecting edges (of which we have E in number), those that bisect the a sides of a face of P all go through a vertex of P0. Thus, exactly a edges meet at a vertex of P0. Finally, those edges of P0 that bisect the b edges at a vertex of Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

222 17. The Five Platonic Solids

ee2 1

m1 m2

m m e3 3 b eb

Figure 17.9

P are the edges of a face of P0. Thus each face of P0 is a regular b- sided polygon. We obtain that the Schli¨ symbol of P0 is {b, a}.We can now study Color Plates 5a–c, which depict the three reciprocal pairs. Notice that the reciprocal of a tetrahedron is another tetrahedron. The reciprocal pair of tetrahedra—the stella octangula,as Kepler called it—is the simplest example of a compound polyhedron, and it occurs in nature as a crystal-twin of tetrahedrite. Histori- cally, the ﬁrst complete understanding of the relationship between a polyhedron and its reciprocal is attributed to Maurolycus (1494– 1575), although inscribing various regular solids into each other appears in Book XV of the Elements. (Note that Book XIV was written by Hypsicles and, as the language and style suggest, Book XV is attributed to several authors.) Reciprocity is a powerful tool in our hands, and we will use it in a variety of ways. First of all, if a regular polyhedron exists, then so does its reciprocal. Thus, the obvious existence of the cube, with Schli¨ symbol {4, 3}, implies the existence of its reciprocal, the octahedron. Moreover, since the dodecahedron and icosahedron are reciprocal, it is enough to show that one of them exists! Since the existence of a tetrahedron is quite clear, our task is now reduced to showing that the dodecahedron exists. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 223

l l2

Figure 17.10

The dodecahedron has Schli¨ symbol {5, 3}, and this shows that exactly three regular pentagons meet at a vertex. This configuration exists (Figure 17.10), since between two adjacent edges of a regular pentagon the angle is π − 2π/5 3π/5, and three of these add up to 9π/5 < 2π. We claim that the lines l1,l2 shown in Figure 17.10 are perpendicular. Indeed, l1 is parallel to the side l, and l is certainly perpendicular to l2. Thus we are able to pick perpendicular diagonal lines in adjacent pentagonal faces. The dodecahedron has exactly twelve faces, so we can pick twelve lines with orthogonality conditions among them. But twelve is exactly the number of edges of a cube! It is impossible to resist the temptation to pick these lines to form the edges of a cube. Figure 17.11 shows the configuration. Notice that we have not proved the existence of the dodecahedron, but have found a cube on which we want to build it. Thus, we start with a cube, pick a vertex, and arrange the three pentagonal faces to meet at this vertex (Figure 17.12). How to attach the remaining pentagonal faces? As usual, group theory helps us out. In fact, we now apply to the installed three pentagonal faces the symmetries of the cube that are spatial half-turns around the symmetry axes that go through the centroids of pairs of opposite faces (see Figure 17.13). Thus the pentagons fit together, and we created the dodecahedron! (For a different proof of the existence of the dodecahedron, cf. Problem 20.) So now we have shown that all five Platonic solids exist. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

224 17. The Five Platonic Solids

Figure 17.11

Pursuing the analogy with regular polygons, we next work out the symmetry groups of these regular polyhedra. Reciprocity reduces the cases to consider to three, since reciprocal polyhedra have the same symmetry group. This follows immediately if we recall how the reciprocal P0 of P was constructed. (Assume that S ∈ Symm (P), construct P0, and realize that S carries P0 to itself because P0 is determined by P using data such as midpoints of edges, etc. that remain invariant under S.) ♥ Before we actually determine these three groups explicitly, we would like to see what the possibilities are for any such group. As noted above, the symmetry group of any polyhedron (regular or not) is ﬁnite, so we now take up the more ambitious task of

Figure 17.12 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 225

Figure 17.13 classifying all finite subgroups of Iso (R3) (cf. Theorem 4 of Section 9 for the 2-dimensional case). Let G ⊂ Iso (R3) be a finite group. First we observe the existence 3 of a point p0 ∈ R that is left fixed by every element of G. Indeed, let p ∈ R3 be arbitrary and consider the orbit G(p) {S(p) | S ∈ G}.

This is a ﬁnite set

G(p) {p1,...,pn} since G is finite. Since the elements of G permute the pj’s, j 1,...,n, we can repeat the argument that was used to find the centroid of a polyhedron and conclude that p +···+p p 1 n 0 n is left fixed by all elements of G. It is quite remarkable that the “centroid argument” for polyhedra applies to this more general situation. We now assume that G consists of direct isometries only. Every nontrivial element R of G is a rotation (see Theorem 10 of Section 16) around an axis l that must go through p0, since R fixes p0. The 2 axis cuts the unit sphere S (p0) around p0 into a pair of antipodal points. We call these points poles (Figure 17.14). Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

226 17. The Five Platonic Solids

2 S ( p 0 )

Figure 17.14

2 There are two poles on S (p0) for each rotation in G. Any rotation R in G must have finite order, since G is finite. We now let q be a pole and R be the smallest positive angle rotation with pole q.We call the order of R the degree of the pole q. Thus, q has degree d ∈ N if Rd I, and d is the smallest positive integer with this property. 2 Let Q denote the (finite) set of all poles on S (p0). We claim that 2 G leaves Q invariant. G certainly leaves S (p0) invariant, since it fixes p0. Let q ∈ Q as above and assume that q corresponds to the rotation R ∈ G. Let S ∈ G be any rotation. Then S ◦ R ◦ S−1 ∈ G is a rotation that fixes S(q) so that it must be a pole. Invariance of Q under G follows. Notice that the poles q and S(q) have the same degree, since Rd I iff (S ◦ R ◦ S−1)d I. The classification of possible cases for G now depends on the successful enumeration of the elements in Q . To do this, we introduce an equivalence relation ∼ on Q such that q1 ∼ q2, q1,q2 ∈ Q , if q2 S(q1) for some S ∈ G. That this is an equivalence follows from the fact that G is a group. The equivalence classes are actually the orbits of G on Q , and they split Q into mutually disjoint subsets. As noted above, poles in the same equivalence class have the same degree. Let C ⊂ Q be an equivalence class and d dC Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 227 the common degree of the poles in C. We claim that |G| |C| . dC To show this, let q ∈ C, and let R ∈ G be the rotation that corresponds to the pole q as above. By assumption, R has degree d,so that the cyclic subgroup 'R( consists of the distinct elements I, R, R2,...,Rd−1 (Rd I).

2 Let p ∈ S (p0) be any point not in Q . Apply the elements in 'R( to p to obtain a regular d-sided polygon with vertices p, R(p), R2(p),...,Rd−1(p).

Other rotations in G transform this polygon into congruent polygons around the poles in C. We can choose p so close to q that all the transformed polygons are disjoint (Figure 17.15). The number of vertices of all these polygons is d|C|. On the other hand, this set of vertices is nothing but the orbit G(p) {S(p) | S ∈ G}.

We obtain that |G(p)| d|C|. Finally, note that no element in G (other than the identity) ﬁxes p, since p is not in Q . Thus, |G(p)| |G|, and the claim follows.

q R(p) R 2 (p)

2 S ( p 0 )

Figure 17.15 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

228 17. The Five Platonic Solids

Next, we count how many nontrivial rotations have poles in C. Each axis, giving two poles in C, is the axis of d − 1 nontrivial rotations. Thus, the number of nontrivial rotations with poles in C is 1 (dC − 1)|G| ,d dC, 2 dC where the (1/2) factor is because each rotation axis gives two poles. Thus, the total number of nontrivial rotations in G is |G| (d − ) |G|− C 1 , 1 d 2 C C where the summation runs through the equivalence classes of poles. Rearranging, we ﬁnd − 2 − 1 . 2 |G| 1 d C C We now see how restrictive this crucial equality is. We may assume that G consists of at least two elements. Since ≤ − 2 < , 1 2 |G| 2 the number of equivalence classes in the summation above can only be 2 or 3. (¬ If we had at least four terms 1 − 1/dC, they would add up to a sum ≥ 4(1 − 1/2) 2. ¬) Assume ﬁrst that we have two equivalence classes, C1 and C2, with dC d1 and dC d2. We have 1 2 2 1 1 2 − 1 − + 1 − |G| d1 d2 or, equivalently, |G| |G| + 2. d1 d2

On the left-hand side the terms are positive integers, since |C1| |G|/d1 and |C2| |G|/d2 as proved above. Thus both terms must be equal to one. We obtain

d1 d2 |G|. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 229

This means that G is cyclic and consists of rotations around a single axis that cuts S1(p0) at an antipodal pair of poles. Toget something less trivial, we assume now that there are three equivalence classes: 2 1 1 1 2 − 1 − + 1 − + 1 − |G| d1 d2 d3 (with obvious notation). We rewrite this as 1 1 1 2 + + 1 + . d1 d2 d3 |G| Since 1/3 + 1/3 + 1/3 1 < 1 + 2/|G|, there must be at least one degree that is equal to 2. We may assume that it is d3. Setting d3 2, the equality above reduces to

1 + 1 1 + 2 . d1 d2 2 |G| A little algebra now shows that d d (d − )(d − ) − 1 2 < . 1 2 2 2 4 1 |G| 4 We obtain the same restriction as for regular polyhedra! Setting, for convenience, d1 ≤ d2, we summarize the possible values of d1, d2, d3, and |G| in the following table:

d1 2333

d2 n 345

d3 2222 |G| 2n 12 24 60

The ﬁrst numerical column d1 2,d2 n, d3 2 and |G| 2n corresponds to the dihedral group Dn discussed in Section 9. There is, however, a little geometric trouble here. Recall that Dn is the symmetry group of a regular n-sided polygon and that this group includes not only rotations but rections as well. Our symmetry Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

230 17. The Five Platonic Solids

group here consists of direct isometries only. This virtual contradiction is easy to resolve. In fact, a spatial half-turn is a direct spatial isometry, yet its restriction to a plane through its axis gives a rection! Thus, increasing the dimension by one enables us to represent our planar opposite isometries by spatial direct isometries. We can now give a “geometric representation” of Dn as follows: Consider 2 1 the regular n-sided polygon Pn ⊂ R inscribed in the unit circle S as in Section 5. Take the Cartesian product of Pn with the interval [−h/2/ 2] ⊂ R, h>0. We obtain what is called a regular prism 2 3 Pn × [−h/2/ 2] ⊂ R × R R (of height h). + We now claim that Symm (Pn × [−h/2/ 2]), the group of direct spatial isometries of√ the prism, is the dihedral group Dn.(Forn 4, we assume that h 2 so that Pn ×[−h/2/ 2] is not a cube.) Since the centroid (the origin) of the prism must stay fixed, it is clear that every element S ∈ Symm (Pn × [−h/2/ 2]) leaves the middle slice 2 Pn ×{0} invariant. Thus, S, restricted to the coordinate plane R spanned by the first two axes, is an element of Symm (Pn). Recall from Section 9 that the elements of Symm (Pn) are re- flections to the symmetry axes or rotations by angles 2kπ/n, k 0,...,n − 1. If S restricts to a planar rection to a symmetry axis of Pn, then S is a spatial half-turn around the same axis. If S restricts to a planar rotation around the origin with angle 2kπ/n, k 0,...,n − 1, then S is a spatial rotation around the vertical third coordinate axis with the same angle. Altogether, we have n + 1 rotation axes, n half-turn axes in R2, and the third coordinate axis (for rotations with angles 2kπ/n, k 0,...,n− 1). These give 2(n+ 1) 2n + 2 poles on the unit sphere S2. The set of poles splits into three equivalence classes. Each half-turn switches the North and South Poles, and these form a two-element equivalence class. On the plane of Pn, there are two intertwining equivalence classes, with n elements in each class. Summarizing, we showed that the dihedral group Dn of spatial rotations is the symmetry group of a 3 regular prism with base Pn. ♠ Conversely, if G acts on R with direct isometries and orbit structure described by the first numerical 3 column of the table above, then G is conjugate to Dn in Iso (R ). Indeed, G contains a degree-n rotation whose powers form a cyclic subgroup of order n in G. The rest of G is made up by n half-turns. The axis of the degree-n rotation is perpendicular to the axes of Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 231 the half-turns. This is because the two poles that correspond to the axis of the degree-n rotation form a single orbit, and these two poles must be interchanged by each half-turn. The 2n poles corresponding to the n half-turns are divided into two orbits consisting of n poles each. The degree-n rotation maps this set of poles into itself. The only way this is possible is that the angle between adjacent axes of the n half-turns is π/n. By a spatial isometry, the configuration of all the axes can be brought to that of the prism above. The same isometry conjugates G into Dn. ♥ This completely describes the first numerical column in the table above. The trivial case of a cyclic G with two equivalence classes, discussed above, can be geometrically represented as the symmetry group of a regular pyramid with base Pn. We could go on and do the same analysis for the remaining columns of the table. Each case corresponds to a single group whose generators and defining relations can be written down explicitly. We follow here a more geometric path. We go back to our five Platonic solids and work out the three symmetry groups that arise. We will then realize (repeating the counting argument above for nearby poles p and q) that they must correspond to the last three columns of the table. We start with the symmetry group of the regular tetrahedron T . Looking at Figure 17.16, we see that the only spatial rections that leave T invariant are those in planes that join an edge to the midpoint of the opposite edge. There are exactly six of these planes,

Figure 17.16 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

232 17. The Five Platonic Solids

each corresponding to an edge of T . Symm (T ) thus contains six rections. Let 1, 2, 3, 4 denote the vertices of T . Each element S ∈ Symm (T ) permutes these vertices; that is, S gives rise to a permutation 1234. S(1)S(2)S(3)S(4)

Conversely, S is uniquely determined by this permutation. (In- deed, if S1 and S2 in Symm (T ) give the same permutation on the S−1 ◦ S vertices, then 2 1 fixes all the vertices. A spatial isometry that S−1 ◦ S I fixes four noncoplanar points is the identity. Thus 2 1 , and S1 S2 follows.) Each rection gives a transposition (a permutation that switches two numbers and keeps the rest of the numbers fixed). In fact, the two vertices that are on the plane of the rection stay fixed, and the other two get switched. The six rections thus 8 give six transpositions. The symmetric group S4 on four letters, consisting of all permutations of {1, 2, 3, 4}, has exactly six transpositions, so we got them all! It is a standard fact (easily verified in our case) that the transpositions generate the symmetry group. ∼ Thus, Symm (T ) S4. The direct isometries in Symm (T ) are compositions of even number of rections. They correspond to permutations that can be written as products of even number of transpositions. These are called even permutations, and they form a subgroup A4 ⊂ S4 called the alternating group on four letters. We obtain that the + group of direct isometries Symm (T ) is isomorphic with A4. Since |S4| 4! 24, we have |A4| |S4|/2 12. Looking at our table, we see that we recovered the second numerical column. The group + ∼ Symm (T ) A4 is called the tetrahedral group. Now explore some rotations in Symm+(T ) as shown in Figures 17.17 to 17.19.

Remark. While the tetrahedral group consists only of rotations, an opposite spatial symmetry of T is not necessarily a rection. For example,

8See “Groups” in Appendix B. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 233

Figure 17.17

Figure 17.18

Figure 17.19 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

234 17. The Five Platonic Solids

the cycle 1234 2341

corresponds to a rotatory reflection (a rotation followed by a rection). Next, we have a choice between a cube and an octahedron, since they are reciprocal to each other. Let us choose the latter. We think of the regular octahedron O in the following way: First take a regular tetrahedron T . At each of the four vertices, we take the vertex figure, which in this case is a triangle whose vertices are the midpoints of the three edges that meet at the given vertex of T .Wenow slice off the four tetrahedra containing the vertices of T along the vertex figures. What is left after this truncation is a regular octahedron O (see Figure 17.20). You may already have noticed in Color Plate 5a that O is actually the intersection of T and its reciprocal T 0. Symmetries of T automatically become symmetries of O: Symm (T ) ⊂ Symm (O). There are exactly four faces of the octahedron O that are contained in those of T . These four faces are nonadjacent and meet only at vertices, and any member determines the group uniquely. The

Figure 17.20 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 235 other four faces of O form the same conﬁguration relative to the reciprocal tetrahedron T 0. Now let S ∈ Symm+(O). We claim that either S(T ) T or S(T ) T 0, depending on whether S is in Symm+(T ) or not. Indeed, looking at how S acts on the two groups of four nonadjacent faces above, we see that S either permutes the faces in each group separately or it interchanges the faces between the two groups. Extending the faces in each group, we obtain T and T 0, and the claim follows (see Color Plate 6). Note that the second case does occur; e.g., take a spatial quarter-turn about an axis that joins two opposite vertices of O. Considering the four possible cases of compositions of two elements of Symm+(O),itis now clear that there are exactly two left-cosets in Symm+(O) by the subgroup Symm+(T ) (corresponding to the two cases above). + + Comparing Symm (T ) ⊂ Symm (O) and A4 ⊂ S4, we see that + Symm (O) is isomorphic to the symmetric group S4.Amoreex- plicit way to obtain this isomorphism is to mark the vertices of one tetrahedron by 1, 2, 3, 4, and their antipodals by 1’, 2’, 3’, 4’ (the vertices of the reciprocal) and to consider the action of Symm+(O) on the “diagonals” 11’, 22’, 33’, 44’.Since |S4| 24, we recovered the third numerical column in our table! The group of direct isometries + ∼ Symm (O) S4 is called the octahedral group.

Remark. Our argument was based on the fact that a regular tetrahedron and its reciprocal intersect in an octahedron. You may be wondering what we get if we intersect the other two reciprocal pairs. Well, we obtain nonregular polyhedra! The intersection of the reciprocal pair of a cube and an octahedron gives what is called a cuboctahedron, a convex polygon with eight equilateral triangular faces and six square faces (Figure 17.21). The intersection of the reciprocal pair of an icosahedron and a dodecahedron is an icosidodecahedron. It has twenty equilateral triangular faces and twelve pentagonal faces (Figure 17.22).

Finally, we take up the task of determining the group of direct isometries of the dodecahedron. It will be easier to work with its reciprocal, the icosahedron I. In Color Plate 7, the twenty faces of Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

236 17. The Five Platonic Solids

Figure 17.21

I have been colored with five different colors, with the property that in each color group the four faces are mutually disjoint. Figure 17.23 shows one color group in a wireframe setting. A simple algorithm to find four faces in a color group is the following: Stand on a face F with bounding edges ek, k 1, 2, 3. For each k, step on the face Fk adjacent to F across ek. At the vertex vk of Fk opposite to ek, five faces meet, three of which are not disjoint from F. The remaining two faces disjoint from F are adjacent and appear to you to the right and to the left. Now, if you are right- handed, add the right face to the color group of F, and if you are left-handed, add the left face to the color group. The plane extensions of each of the four faces in a color group enclose a regular tetrahedron (Figure 17.24). Since each color

Figure 17.22 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 237

Figure 17.23 group gives one tetrahedron, altogether we have five tetrahedra. The five tetrahedra is one of the most beautiful compound polyhedra (Figure 17.25). Its Schli¨ symbol {5, 3}[5{3, 3}]{3, 5} rects that the twenty vertices of the five tetrahedra give the vertices of a dodecahedron (see Problem 19), and the twenty faces enclose an icosahedron (see the front cover illustration). (With this terminology, the reciprocal pair of two tetrahedra has Schli¨ symbol {4, 3}[2{3, 3}]{3, 4} (see Problem 18).)

Figure 17.24 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

238 17. The Five Platonic Solids

Figure 17.25

Now let S ∈ Symm+(I). Looking at the way we colored the icosahedron, we see that S acts on the set of the five tetrahedra as an even permutation, so Symm+(I) can be represented as a subgroup of A5. The twelve spatial rotations of a fixed tetrahedron act on the remaining four tetrahedra as even permutations. We obtain that the group of direct isometries Symm+(I) is isomorphic with the alternating group A5 on five letters. We arrive at the icosahedral group.

Remark. ♠ The icosahedral group plays a prominent role in Galois theory in connection with the problem of solving quintic (degree 5) polynomial equations in terms of radicals (in a similar way as the quadratic formula solves all quadratic equations). In fact, Galois proved that a polynomial equation is solvable by radicals iff the associated Galois group (the group of automorphisms of the splitting ﬁeld) is solvable. The connection beween the symmetries of the icosahedron and unsolvable quintics is subtle9 but it is based on the fact that a quintic with Galois group A5 is an example of a degree 5 equation for which no root formula exists. Any irreducible (over Q ) quintic with exactly 3 real roots (and a pair of complex conjugate roots) provides such an example.

9See F. Klein, Lectures on the icosahedron, and the solution of equations of the ﬁfth degree,Trubner¨ and Co., 1888. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 239

Remark. ♣ Our understanding of the octahedral and icosahedral groups was based on their tetrahedral subgroups. Many different treatments of this topic exist. For example, given a Platonic solid P with Schli¨ symbol {a, b}, it is clear that the axis of a rotational symmetry must go through a vertex or the midpoint of an edge or the centroid of a face. In the ﬁrst case, the rotation angle is an integer multiple of 2π/b; in the second, an integer multiple of π; in the third, an integer multiple of 2π/a. The number of nontrivial rotations is therefore (1/2)(V(b − 1) + E + F(a − 1)), where V, E, and F are the number of vertices, edges, and faces of P. The one-half factor is present because when considering lines through vertices, midpoints of edges, and centroids of faces, we actually count the rotation axes twice. Since 2E aF, this number is (1/2)(2E + 2E − 2) 2E − 1, where we also used Euler’s theorem. Thus the order of Symm+(P) is 2E.

We summarize our hard work in the following:10

Theorem 11. The only ﬁnite groups of direct spatial isometries are the cyclic groups Cn, the dihedral groups Dn, the tetrahedral group A4, the octahedral group S4, and the icosahedral group A5.

The argument in the remark after the list of the 17 crystallographic groups in Section 10 can be adapted to spherical geometry to give another proof of Theorem 11. We assume that G isaﬁ- 3 nite group of rotations in R with ﬁxed point p0 and set of poles 2 Q ⊂ S (p0).IfQ contains only one pair of antipodal points, then G is cyclic. From now on we assume that this is not the case. Let R2α(p) and R2β(q) be rotations in G with least positive angles 2α and 2β and distinct poles ±p and ±q. Since α and β are minimal, π/α and π/β are integers (≥ 2). Following the argument in Section 10 cited above, we obtain another pole r ∈ Q such that p, q, r are

10 This result is due to Klein. Note also that, based on analogy with Euclidean plane isometries, a possible continuation of this topic would include discrete subgroups of Iso (R3) and 3-dimensional crystallography. Alas, we are not prepared to explore this beautiful subject here. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

240 17. The Five Platonic Solids

2 vertices of a spherical triangle in S (p0) with angles α, β, γ, and we have

R2α(p) ◦ R2β(q) ◦ R2γ (r) I.

In particular, if α β π/2 (that is, R2α(p) and R2β(q) are half- turns), then r is perpendicular to the great circle through ±p and ±q. Then the rotation R2γ is a half-turn iff p and q are perpendicular. We see that if a noncyclic ﬁnite group G consists of half-turns only, then G must be isomorphic to the dihedral group D2 of order 4, containing exactly three half-turns with mutually perpendicular axes. In a similar vein, if G contains exactly one pair of poles of degree greater than or equal to 3, then all the poles corresponding to the half-turns in G must be perpendicular to this, and G must be dihedral. From now on we assume that G contains at least two rotations of degrees greater than or equal to 3 and distinct axes. Using the notation above, we choose R2α(p) and R2β(q), two rotations in G of degrees greater than or equal to 3, and distinct poles ±p and ±q such that the (spherical) distance between p and q is minimal among all the poles in Q of degree greater than or equal to 3. As above, we have R2γ (r) R−2β(q) ◦ R−2α(p) ∈ G. By the minimal choice of p, q, in addition to π/α and π/β being integral, π/γ must also be an integer greater than or equal to 3. The spherical excess formula (cf. Section 13) gives α β γ + + > . π π π 1 In particular, at least one of the integers π/α, π/β, π/γ must be 2. Because of our initial choices, π/α, π/β ≥ 3, we must have π/γ 2. The constraint above reduces to α β + > 1 . π π 2 Rearranging, we obtain (π/α − 1)(π/β − 2)<4, π/α, π/β ≥ 3. Since π/α and π/β are integers, one of them must be 3 and the other must be 3, 4, or 5. A simple analysis shows that the subgroup G0 ⊂ G generated by R2α(p) and R2β(q) is the symmetry group of a tetrahedron, an octahedron/cube, or an icosahedron/dodecahedron. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 241

More precisely, the orbit G0(p) is the set of vertices of a regular polygon with Schli¨ symbol {π/β, π/α}. In fact, the powers of R2β(q) applied to p give a (π/β)-gonal face, and the powers of R2α(p) transform this face to the configuration of π/α faces sur- rounding the vertex p. The rest of G0 installs the remaining faces of the polyhedron with vertices G0(p). In a similar vein, G0(q) is the set of vertices of the reciprocal polyhedron with Schli¨ symbol {π/α, π/β}, while G0(r) is the set of common midpoints of the edges of both polyhedra. For convenience, we now project these 2 polyhedra to S (p0) radially from p0, and obtain spherical tessellations. The spherical triangle with vertices p, q, r is characteristic in the sense that in a flag of the spherical polyhedron with vertices G0(p), p is a vertex, q is the midpoint of a face, and r is the midpoint of an edge in the flag. More about this in Sections 24–25. Now, it takes only a moment to realize that G0 G. Indeed, by the minimal choice of p, q, G − G0 cannot contain any rotations of order greater than or equal to 3, since one of the corresponding poles of degree greater than or equal to 3 (transformed by an appropriate element in G0) would show up in the face of the spherical polygon with centroid q and vertices in G(p). Thus G − G0 can contain only half-turns. If Rπ(s) were a half-turn in G − G0, then Rπ(r) ◦ Rπ(s) ∈ G − G0 would also be a half-turn, so that r and s would be perpendicular. Thus the axis R ·s would be perpendicular to all axes R · g0(r), g0 ∈ G0, and this is impossible. Thus G G0, and Theorem 11 follows. We also realize that we have obtained the following as a byproduct: Any finite group of rotations in R3 is generated by one or two elements!

Remark. ♠ We now pick up the opposite isometries. Let G ⊂ Iso (R3) be any ﬁnite group. Let G+ ⊂ G denote the subgroup consisting of the direct isometries of G. The possible choices of G+ are listed in the theorem above. To get something new, we may assume that G+ ⊂ G is a proper subgroup. Since composition of two opposite isometries is direct, G+ is of index 2 in G, that is, |G| 2|G+|.In other words, G+ and any element in G− G − G+ generate G.We now have to study the possible conﬁgurations of G+ (listed above) and a single opposite isometry. We have two cases, depending on Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

242 17. The Five Platonic Solids

whether the antipodal map −I : R3 → R3 (which is opposite in R3) belongs to G or not. If −I ∈ G, then G− (−I) · G+, since the right-hand side is a set of |G+| opposite isometries in G. Since −I commutes with the elements of G, the cyclic subgroup {±I}⊂G (isomorphic to C2) is normal. Since index-2 subgroups + ∼ + (such as G ⊂ G) are always normal, we obtain G G × C2. Using Theorem 11, we arrive at the list

Cn × C2,Dn × C2, A4 × C2, S4 × C2, A5 × C2. Assume now that −I ∈ G. We first show that G+ is contained in a (finite) group G∗ of direct isometries as an index-2 subgroup. To do this, we define G∗ G+ ∪ (−I) · G−.

Notice that G∗ is a group, since −I commutes with the elements of G. Moreover, G∗ consists of direct isometries, since −I and the elements of G− are opposite. G+ and (−I) · G− are disjoint, since −I ∈ G. In particular, |G∗| 2|G+|, and G+ is an index-2 subgroup in G∗. The possible inclusions G+ ⊂ G∗ are easily listed, since all the isometries involved are direct and Theorem 11 applies. We obtain

Cn ⊂ C2n,Cn ⊂ Dn,Dn ⊂ D2n, A4 ⊂ S4. Finally, G can be recovered from G∗ via the formula G G+ ∪ (−I) · (G∗ − G+). In general, we denote by G∗G+ the group defined by the right-hand side of this equality, when G+ ⊂ G∗ is an inclusion of finite groups of direct isometries, and G+ is of index 2 in G∗. We finally arrive at the list

C2n · Cn,Dn · Cn,D2n · Dn, S4 · A4. For example, for the full symmetry groups of the Platonic solids, we have the following: + + Symm (T ) Symm (O) · Symm (T ) S4 · A4, + Symm (O) Symm (O) × C2 S4 × C2, + Symm (I) Symm (I) × C2 A5 × C2. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 243

Figure 17.26

♣ Aside from constructibility, for computational purposes it will be important to realize the five Platonic solids in convenient positions in R3. Since every convex polyhedron is uniquely determined by its vertices, our task is to find for each Platonic solid a “symmetric” position with vertex coordinates as simple as possible. We start with the tetrahedron T . A natural positon of T in R3 is defined by letting the coordinate axes pass through the midpoints of three edges, as shown in Figure 17.26. The four vertices of T are

(1, 1, 1), (1, −1, −1), (−1, 1, −1), (−1, −1, 1).

The tetrahedral group contains the three half-turns around the coordinate axes and the rotation around the “front vertex” (1, 1, 1) by angle 2π/3. It is easy to see that these four rotations generate the twelve rotations that make up the tetrahedral group. It would be very easy to write down these transformations in terms of orthogonal 3 × 3 matrices. We skip this, since the description of Symm+(T ) is much easier using quaternions, which we will discuss in Section 23. Truncating the tetrahedron by chopping off the four tetrahedra along the vertex ﬁgures, we arrive at the octahedron O with vertices

(±1, 0, 0), (0, ±1, 0), (0, 0, ±1).

Taking the midpoints of the edges of O, we obtain the midpoints of the reciprocal cube C: ± 1 , ± 1 , ± 1 . 2 2 2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

244 17. The Five Platonic Solids

The cube described this way has edge length 1, and the edges are parallel to the coordinate axes. The centroid of all three Platonic solids T , O, and C is the origin. To realize the dodecahedron D and the icosahedron I in R3 is a less trivial task. We start with the icosahedron I and make the following observation. Let v be a vertex of I. The ﬁve faces of I that contain v form a pyramid whose base is a regular pentagon. We call v the vertex of the pentagonal pyramid. Taking the edge length of I to be 1, the length τ of a diagonal of the regular pentagonal base is the so-called golden section (attributed to Eudoxus in 400 b.c.) π π π τ 2 sin − 2 cos 2 5 5 (Figure 17.27). Inserting two additional diagonals in the base, an interesting picture emerges (Figure 17.28). Notice that p0,p1,p2,p3 is a rhombus and the triangles %p0p1p3 and %p0p4p5 are similar. Thus, d(p0,p4) 1/τ. But adding 1 to this gives the diagonal again: 1 + 1/τ τ. Multiplying out, we see that τ is the positive solution of the quadratic equation

τ2 − τ − 1 0.

Figure 17.27 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 245

τ p1 p3

1/τ

p5 p4 Figure 17.28

The quadratic formula gives √ + τ 5 1 . 2

Remark. Alternatively, the golden section τ can be deﬁned as the unique ratio of side lengths of a rectangle with the property that if a square is sliced off, the remaining rectangle is similar to the original rectangle. This deﬁnition gives τ − 1 1/τ, which leads to the same quadratic equaion as above. Used recursively, a circular quadrant can be inserted into each sliced off square to obtain an approximation of an Archimedes spiral (see Figure 17.29). The most commonly known phenomenon in nature that patterns this is

Figure 17.29 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

246 17. The Five Platonic Solids

the nautilus shell, where the shell grows in a spiral for structural harmony in weight and strength. Iterating τ 1 + 1/τ, we obtain the continued fraction 1 τ 1 + , + 1 1 1+··· and this can be used to give rational approximations√ of τ (cf. Prob- lem 14). In a similar vein, iterating τ 1 + τ, we arrive at the inﬁnite radical expansion √ τ 1 + 1 + 1 +···.

Recall that the icosahedron I can be sliced into two pentagonal pyramids with opposite vertices v1 and v2 and a pentagonal antiprism (a belt of ten equilateral triangular faces). Consider now two opposite edges e1 and e2 of the pyramids emanating from v1 and v2. They are parallel and form the opposite sides of a rectangle whose longer sides are diagonals of two pentagonal bases! Since the side lengths of this rectangle have ratio τ : 1, it is called a golden rectangle. (Ancient Greeks attributed special significance to this; for example, the Parthenon in Athens (fifth century b.c.) fits perfectly in a golden rectangle.) A beautiful model of the icosahedron (due to Pacioli and shown in Figure 17.30) emerges this way; the twelve vertices of I are on three golden rectangles in mutually perpendicular planes! Considering now these planes as coordinate planes for our coordinate system, we see immediately that the

Figure 17.30 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

17. The Five Platonic Solids 247

Figure 17.31 twelve vertices of the icosahedron are (0, ±τ, ±1), (±1, 0, ±τ), (±τ, ±1, 0). A golden rectangle can be inscribed in a square such that each vertex of the golden rectangle divides a side of the square in the ratio τ : 1 (Figure 17.31). Inserting these squares around the three golden rectangles that make up the icosahedron, we obtain an octahedron circumscribed about I (Figure 17.32). Looking at Figure 17.33, we see that the vertices of the octahedron are (±τ2, 0, 0), (0, ±τ2, 0), (0, 0, ±τ2) and this is homothetic (with ratio of magniﬁcation τ2) to our earlier model O.

Figure 17.32 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

248 17. The Five Platonic Solids

Figure 17.33 τ

Finally, note that the dodecahedron D constructed on the cube with vertices (±1, ±1, ±1) as above has vertices

(0, ±1/τ, ±τ), (±τ, 0, ±1/τ), (±1/τ, ±τ, 0).

That these are the correct vertices for D can be veriﬁed using reciprocity. An alternative approach is given in Problem 20.

Problems

∼ 1. Using the isomorphism Symm(T ) S4, work out for the regular tetrahedron the rotational symmetries that correspond to all possible products of two transpositions.

2. Let T be a regular tetrahedron. (a) Given a flag (p,e,f) of T , what is the composition of S1,S2 ∈ + Symm (T ), where S1 and S2 are counterclockwise rotations by 2π/3 about p and the centroid of f ? (b) Given opposite edges e1 and e2, list all possible scenarios for flags (p1,e1,f1) and (p2,e2,f2) and symmetries of T that carry one flag to the other. (c) Let e1 and e2 be opposite edges. Show that the midpoints of the four complementary edges are vertices of a square. Slice T with the plane spanned by this square; prove that the two pieces are congruent and find a symmetry of T that carries one piece to the other. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 249

3. Let S ∈ Symm+(O) be an isometry that does not leave T invariant. Show that S ◦ Symm+(T ) ◦ S−1 Symm+(T ). + 4. Derive that Symm (O) is isomorphic to S4 using a cube rather than an octahedron. 5. (a) Construct a golden rectangle and a regular pentagon with straightedge and compass. (b) Use the fact that the vertices of a regular pentagon inscribed in the unit circle are the powers ωk, k 0,...,4, of the primitive fifth root of unity ω e2πi/5 to show that ω − ω4 τ −(ω2 + ω3) 1 . ω2 − ω3 ω + ω4 (c) Use (b) to derive the formula τ2 + 1 −i(ω − ω4). (d) Let s and d be the side length and the diagonal length of a regular pentagon; τ d/s. Show that the side length and the diagonal length of the regular pentagon whose sides extend to the five diagonals of the original pentagon are 2s − d and d − s. Interpret this via paper folding. Conclude that τ is irrational. (Hint: Assume that τ d/s with s and d integral, and use Fermat’s method of infinite descent. Compare this with Problem 12 in Section 3.) 6. Prove that the icosahedral group Symm+(I), is simple11 using the following argument: Let N ⊂ Symm+(I) be a normal subgroup. (a) Show that if N contains a rotation with axis through a vertex of I then N contains all rotations with axes through the vertices of I. (b) Derive similar statements for rotations with axes through the midpoints of edges and the centroids of faces of I. (c) Counting the nontrivial rotations in N, conclude from (a)–(b) that |N| 1 + 24a + 20b + 15c, where a, b, c are0or1. (d) Use the fact that |N| divides 60 to show that either a b c 0or a b c 1. + 7. Establish an isomorphism between Symm (I) and A5 by filling in the details for the following steps: (a) There are exactly five cubes inscribed in a dodecahedron (see Color Plate 8). (b) The icosahedral group permutes these cubes. + (c) The map φ : Symm (I) → S5 defined by the action in (b) is an injective homomorphism with image A5.

11 See “Groups” in Appendix B. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

250 17. The Five Platonic Solids

8. True or false: Two cubes inscribed in a dodecahedron have a common diagonal. (The five cubes inscribed in a dodecahedron form a compound polyhedron with Schli¨ symbol 2 {5, 3}[5{4, 3}], where the 2 means that each vertex of the dodecahedron belongs to two of the cubes (see the back cover illustration). It can also be obtained from the compound of five tetrahedra by circumscribing a cube around each participating tetrahedron and its reciprocal.) 9. Find two flags in Bimbo’s lozenge that cannot be carried into each other by a symmetry. 10. Classify spherical tilings following the argument in the remark at the end of Section 10. Observe that for a spherical triangle with angles α, β, and γ such that π/α, π/β, and π/γ are integers, the inequality α/π + β/π + γ/π > 1 gives only finitely many possibilities. 11. Show that the area of a spherical n-sided polygon is the sum of its angles minus (n − 2)π. 12. Prove Euler’s theorem for convex polyhedra using Problem 11 as follows: Let K be a convex polyhedron. Place K inside S2 (by scaling if necessary) such that K contains the origin in its interior. Project the boundary of K onto S2 from the origin. Sum up all angles of the projected spherical graph with V vertices, E edges, and F faces in two ways: First, counting the angles at each vertex, find that this sum is 2 . Second, count the angles for each face by converting the angle sum into spherical area (Problem 11) and use that the total area of S2 is 4π. 13. Show that a polyhedron is regular iff its faces and vertex figures are regular.

12 14. ♦ Let Fn denote the nth Fibonacci number deﬁned recursively by F1 1, F2 1, and Fn+2 Fn+1 +Fn, n ∈ N.(Fn can be interpreted as the number of offspring generated by a pair of rabbits, assuming that the newborns mature in one iteration and no rabbits die.) (a) Let qn Fn+1/Fn. Show that the nth convergent of the continued fraction 1 τ 1 + , + 1 1 1+···

is qn. (b) Prove that limn→∞ qn τ. (This observation is due to Kepler.) Use induction with respect to n to show that the ratios qn alternate above and below τ, that is, qn <τfor n odd, and qn >τfor n even. (c) Verify the n+1 recurrence relation τ Fn+1τ + Fn. Deﬁne F−n F−n+2 − F−n+1 for n+ nonnegative integers n, so that F−n (−1) 1Fn, and extend the validity of the recurrence relation to all integers n. (d) Deﬁne the nth Lucas number Ln recursively by Ln+1 Ln + Ln−1, n ∈ N, L0 2, L1 1. Show that Ln

12Leonardo of Pisa (13th century) wrote under the name Fibonacci, a shortened version of ﬁlius Bonacci (son of Bonacci). Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 251

F√n−1 + Fn+1, n ≥ 0. Verify that limn→∞ Ln+1/Ln τ and limn→∞ Ln/Fn n n n 5. Use (c) to show that Ln τ + (−1) /τ . Derive the formula τn − (−1)n/τn Fn , τ + 1/τ due to Binet (1843). 15. (a) Show that the sum of vectors from the centroid to each vertex of a Platonic solid is zero. (b) Prove that the midpoint of each edge of a Platonic solid is on a sphere. (c) Derive an analogous statement for the centroids of the faces. (d) Show that the centroids of the faces of a Platonic solid are the vertices of another Platonic solid. What is the relation between these two Platonic solids? 16. Show that the icosahedron can be truncated13 such that the resulting convex polyhedron has 12 pentagonal and 20 hexagonal faces. This polyhedron is called the buckyball (Figure 17.34). (It has great signiﬁcance in chemistry, since, besides graphite and diamond, it is a third form of pure carbon (Figure 17.35).) Show that the buckyball14 has the following rotational symmetries: (a) Half-turns around axes that bisect opposite pairs of edges; (b) Rotations around axes through the centroids of opposite pairs of hexagonal faces with rotation angles that are integral multiples of 2π/3; (c) Rotations around axes through the centroids of opposite pairs of pentagonal faces, with rotation angles that are integral multiples of 2π/5. (d) Realize that (a)–(b)–(c) account for 59 nontrivial rotations, so that (adding the identity) we obtain the icosahedral group. (e) Show that no vertex stays ﬁxed under any nontrivial rotational symmetry of the buckyball.

Figure 17.34

13Truncating the ﬁve Platonic solids in various ways, one arrives at the thirteen Archimedean solids. It is very probable that Archimedes knew about these, but the ﬁrst surviving written record is by Kepler from 1619. We already encountered two of these, the cuboctahedron and the icosidodecahedron. 14For an excellent article, see F.Chung and S. Sternberg, Mathematics and the Buckyball, American Scientist, Vol. 81 (1993). Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

252 17. The Five Platonic Solids

Figure 17.35

17. (a) Show that the reciprocal of the side length of a regular decagon inscribed in the unit circle is the golden section. (b) Inscribe an equilateral triangle into a circle. Take a segment connecting the midpoints of two sides of the triangle and extend it to a chord of the circle. Show that one midpoint splits the length of the chord in the square of the golden section. 18. The 8 vertices of a reciprocal pair of tetrahedra are those of a (circumscribed) cube (cf. Problem 4). Show that this cube is homothetic to the reciprocal cube of the octahedral intersection of the two tetrahedra. What is the ratio of magniﬁcation? 19. Show that the 20 vertices of the 5 tetrahedra circumscribed to an icosahedron are the vertices of a dodecahedron. Prove that this dodecahedron is homothetic to the dodecahedron reciprocal to the icosahedron. 20. A diagonal splits a pentagon with unit side length into an isosceles triangle and a symmetric trapezoid with a base length of the golden section. Deﬁne a convex polyhedron with a square base of side length that of the golden section, and let it have four additional faces: two isosceles triangles and two symmetric trapezoids, attached to each other along their unit-length sides in an alternating manner. We call this polyhedron a roof.15 The top unit-length side of the trapezoids, which forms a single edge opposite the base, is called the ridge. Notice that the solid dodecahedron is the union of an inscribed cube and six roofs whose bases are the faces of the cube (Figure 17.11). (a) Prove the existence of the dodecahedron by showing that in a roof the dihedral angle between a triangular face and the base and the dihedral angle between a trapezoidal face and the base are complementary. (b) In a given roof, extend the lateral sides of a trapezoid beyond the ridge to form an isosceles triangle. Prove that the four nonbase vertices of the four

15The four planes spanned by the four faces of a tetrahedron divide R3 into the tetrahedron itself, four frusta, four trihedra, and six roofs. See M. Berger, Geometry I–II, Springer, 1980. Although this is unbounded, we use this classical terminology in our situation. The author’s students called the argument for the existence of the dodecahedron the roof-proof. It is essentially contained in Book XIII of the Elements. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 253

isosceles triangles obtained from four trapezoids in an opposite pair of roofs form a golden rectangle (see Color Plate 9). (c) Show that two edges of the golden rectangle in (b) contain the ridges of another pair of opposite roofs. (d) By (b), the three pairs of opposite roofs in a dodecahedron give three mutually perpendicular golden rectangles around which an icosahedron can be circumscribed. Prove that this icosahedron is homothetic to the reciprocal of the dodecahedron. What is the ratio of magniﬁcation? (e) Circumscribe an octahedron to the icosahedron obtained in (d) and relate it to the reciprocal of the cube inscribed in the dodecahedron.

21. A triangular array of dots consists of rows of 1, 2,...,ndots, n ∈ N, stacked up in a triangular shape. The total number of dots in a triangular array is the nth triangular number Tri (n) 1 + 2 +···+n n(n + 1)/2 (see Problem 11 of Section 2). A tetrahedral array of dots consists of triangular arrays of 1, 3,...,n(n+ 1)/2 dots stacked up in a tetrahedral shape. The total number (n) n k(k + )/ n of dots Te t k 1 1 2 in a tetrahedral array is called the th tetrahedral number. In a similar vein, define the nth square pyramid number Pyr (n) 12 + 22 +···+n2 as the total number of dots in a square pyramid obtained by stacking up square arrays of 1, 4,...,n2 dots. Finally, define the nth octahedral number16 Oct (n) Pyr (n) + Pyr (n − 1) as the total number of dots in the octahedron viewed as the union of two square pyramids. (a) Show that Pyr (n) n(n + 1)(2n + 1)/6 by fitting six square pyramids in an n × (n + 1) × (2n + 1) box (cf. also Problem 8 of Section 2). (b) Prove that Pyr (n) 2 Te t (n − 1) + n(n + 1)/2 by slicing the array of dots in a square pyramid into two tetrahedral arrays and a triangular array. Conclude that Te t (n) n(n + 1)(n + 2)/6. (This is attributed to the Hindu mathematician Aryabhatta, c. a.d. 500.) (c) Show that Oct (n) Te t (2n − 1) − 4 Te t (n − 1) n(2n2 + 1)/3 by considering the octahedron as a tetrahedron truncated along the vertex figures.

22. Use the 3-dimensional ball model D3 of hyperbolic space geometry (see Prob- lem 7 of Section 13) to prove the existence of a hyperbolic dodecahedron with right dihedral angles (cf. Problem 8 of Section 14). Show that D3 can be tes- sellated by hyperbolic dodecahedra with right dihedral angles. (Observe the close analogy between dodecahedral tessellations of the hyperbolic space and tessellations of the Euclidean space by cubes.)

23. Prove that (a) the full symmetry group of the regular pyramid with base Pn is Dn · Cn, and (b) the full symmetry group of the regular prism with base Pn is Dn × C2 for n even, and D2n · Dn for n odd.

16For an interesting account on these and other related numbers, see J. Conway and R. Guy, The Book of Numbers, Springer, 1996. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

254 17. The Five Platonic Solids Web Sites

1. www.mathsoft.com/asolve/constant/gold/gold.html 2. www.vashti.net/mceinc/golden.htm 3. www.geom.umn.edu/docs/education/institute91/handouts/node6.html 4. www.geom.umn.edu/graphics/pix/Special Topics/Tilings /allmuseumsolids.html 5. www.frontiernet.net/∼imaging/polyh.html

Film

Ch. Gunn and D. Maxwell: Not Knot, Geometry Center, University of Minnesota; Jones and Bartlett Publishers, Inc. (20 Park Plaza, Suite 1435, Boston, MA 02116). Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Finite Mobius¨ . . . . . Groups . 18 . SECTION .

† ♣ Recall from Section 11 that a spherical rection RC in a great circle C ⊂ S2 can be pulled down by the stereographic projection 2 ˆ hN : S → C (extended to the North Pole N by hN (N) ∞)to ˆ ˆ a rection RS : C → C in the circle S hN (C). In other words, −1 RS hN RChN . Utilizing the rich geometric setting, we were able to derive an explicit formula for RS. A spherical rotation Rθ(p0) 2 around an axis R · p0, p0 ∈ S , is the composition of two spherical rections (in great circles that meet in angle θ/2atp0). Conju- gating (this time) Rθ(p0) by the stereographic projection, we thus obtain the composition of two rections in circles intersecting at hN (p0) in an angle of θ/2, a direct Mobius¨ transformation with ﬁxed point hN (p0). In Section 12 we learned that direct Mobius¨ transformations are nothing but linear fractional tranformations of the ˆ extended complex plane C. Thus, to Rθ(p0) there corresponds a linear fractional transformation whose coefﬁcients depend on θ and the coordinates of p0. It is natural to try to determine these dependencies explicitly. This is given in the folowing theorem, due to Cayley in 1879.

255 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

256 18. Finite Mobius¨ Groups

Cayley’s Theorem. 2 For p0 (a0,b0,c0) ∈ S and θ ∈ R, the spherical rotation Rθ(p0) conjugated with the stereographic projection hN is a linear fractional transformation given by λz −¯µ −1 ˆ N ◦ Rθ(p ) ◦ h )(z) ,z∈ C, 0 N µz + λ¯

where θ θ θ λ cos + sin c i and µ sin (b + a i). 2 2 0 2 0 0

Remark. ♠ Under the isomorphism M¨ob(Cˆ ) SL(2, C)/{±I}, the linear fractional transformation in Cayley’s theorem corresponds to the matrix

λ −¯µ A , |λ|2 +|µ|2 ,λ,µ∈ . µ λ¯ 1 C

This is a special unitary matrix, that is,

A∗ A¯ + A−1,

and A has determinant 1. The special unitary matrices form a subgroup of SL(2, C), denoted by SU(2). The importance of this subgroup will be apparent in Sections 22 to 24.

Proof. ♣ The statement is clear for p0 N, since under hN , Rθ(p0) cor- iθ 3 responds to multiplication by e . Let {e1,e2,N}⊂R denote the −1 standard basis. We work out the composition hN Rθ(p0 N in the particular case p0 e1 (1, 0, 0). First note that the matrix of Rθ(e1) is    10 0 R  0 cos(θ) − sin(θ)  . 0 sin(θ) cos(θ) Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

18. Finite Mobius¨ Groups 257

Using this and the explicit form of hN given in Problem 1 of Section 7, we have z |z|2 − −1 2 1 N ◦ Rθ(p ) ◦ h )(z) N ◦ Rθ(p )) , 0 N 0 |z|2 + 1 |z|2 + 1 2!(z) 2"(z) |z|2 − 1 hN , cos(θ) − sin(θ), |z|2 + 1 |z|2 + 1 |z|2 + 1 2"(z) |z|2 − 1 sin(θ) + cos(θ) |z|2 + 1 |z|2 + 1 !(z) + i"(z) (θ) − i(|z|2 − ) (θ) 2 2 cos 1 sin |z|2 + 1 − 2"(z) sin(θ) − (|z|2 − 1) cos(θ) (z (θ/ ) + i (θ/ ))(z¯ (θ/ ) + i (θ/ )) cos 2 sin 2 sin 2 cos 2 (iz sin(θ/2) + cos(θ/2))(z¯ sin(θ/2) + i cos(θ/2)) (θ/ )z + i (θ/ ) cos 2 sin 2 . i sin(θ/2)z + cos(θ/2) The formula follows in this special case. Turning to the general −1 case, we first note that hN Rθ(p0 N is a linear fractional transformation of the form above (with |λ|2 +|µ|2 1), since any rotation can be written as the composition of rotations with axes R · N and R · e1, and these cases have already been treated. Thus, it remains to show that the explicit expressions for the coefficients λ and µ are valid. Todo this we use the result from Section 12 asserting that a linear fractional transformation is uniquely determined by its action on three points. For the first two points we choose the fixed points of the linear fractional transformation claimed to be equal to our conjugated spherical rotation. The fixed points are obtained by solving the quadratic equation

µz2 − 2i"(λ)z +¯µ 0 for z. We ﬁnd that the ﬁxed points are (c0 ±1)/(a0 −b0i) hN (±p0). For the third point we choose ∞, and verify that under hN , 2 2 Rθ(p0)(N) corresponds to λ/µ. Indeed, since |λ| +|µ| 1, we have

−1 2 2 2 hN (λ/µ) (2λµ,¯ |λ| −|µ| ) ∈ S , Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

258 18. Finite Mobius¨ Groups

and this is Rθ(p0)(N), as an easy computation shows. The theorem follows.

Remark. The first part of the proof of Cayley’s theorem can be skipped if we use the fact that hN is conformal and that any conformal ˆ −1 transformation of C (such as hN Rθ(p0 N ) is a linear fractional transformation. To simplify the terminology, we say that the linear fractional transformation with parameters λ, µ, and Rθ(p0) correspond to each other. Notice that replacing the rotation angle θ by θ + 2π has the effect of changing λ and µ to their negatives. Our main quest, to be fully accomplished in Section 24, is to classify all finite subgroups of the Mobius¨ group Mob¨ (Cˆ ). To begin with, here we confine ourselves to giving a list of finite Mobius¨ groups most of which arise from the geometry of Platonic solids. An example to start with is the cyclic group Cn of order n. This group can be realized as the group of rotations

z → e2kπi/nz, k 0,...,n− 1.

Notice that each rotation can be viewed as a linear fractional tran- kπi/n formation with λ e and µ 0, k 0,...,n− 1. In fact, Cn corresponds to the group of rotations R2kπ/n(N), k 0,...,n− 1, with common vertical axis through the North and South Poles. If we adjoin to Cn the linear fractional transformation z → 1/z (characterized by λ 0 and µ i) that corresponds to the half- turn Rπ(e1), we obtain the dihedral Möbius group Dn of order 2n: e−2kπi/n z → e2kπi/nz, ,k ,...,n− . z 0 1 This group corresponds to the symmetry group of Klein’s dihedron, the regular spherical polyhedron with two hemispherical faces, n spherical edges, and n vertices distributed equidistantly along the 2 equator of S . If we fix one vertex at e1, then the half-turn Rπ(e1) is a symmetry of the dihedron that interchanges the two faces. We also see that the first group of linear fractional transformations in the dihedral Mobius¨ group above corresponds to λ ekπi/n, µ 0, Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

18. Finite Mobius¨ Groups 259 k 0,...,n− 1, and the second corresponds to λ 0, µ iekπi/n, k 0,...,n− 1. As a straightforward generalization, we inscribe a Platonic solid P in S2, apply Cayley’s theorem, and obtain a finite Mobius¨ group G isomorphic to the symmetry group of P. Since reciprocal pairs of Platonic solids have the same symmetry group, we may restrict ourselves to the tetrahedron, octahedron, and icosahedron. We call the corresponding groups tetrahedral, octahedral, and icosahedral Möbius groups. The dihedral Mobius¨ group discussed above can be considered as a member of this family if we replace P with its spherical tessellation obtained by projecting P radially from the origin to S2. In what follows we call these configurations spherical Platonic tesselations. Before we actually determine the Mobius¨ groups explicitly, we should note that, again by Cayley’s theorem, the Mobius¨ groups obtained by inscribing the same Platonic solid into S2 in two different ways are conjugate subgroups in Mob¨ (Cˆ ). Thus, we can choose our spherical Platonic tesselations in convenient positions in S2. We choose our regular tetrahedron such that its vertices are alternate vertices of the cube, and the cube is inscribed in S2 such that its faces are orthogonal to the coordinate axes. We also agree that the first octant contains a vertex of the tetrahedron. This is a scaled version of the regular tetrahedron discussed in Section 17. (The scaling is only to circumscribe S2 around the tetrahedron.) The vertex in the first octant must be √1 , √1 , √1 . 3 3 3 The three coordinate axes go through the midpoints of the three opposite pairs of edges of the tetrahedron. The three half-turns around these axes are symmetries of the tetrahedron. Applying these half-turns to the vertex above, we obtain the remaining three vertices √1 , − √1 , − √1 , − √1 , √1 , − √1 , − √1 , − √1 , √1 . 3 3 3 3 3 3 3 3 3

As noted above, the half-turn Rπ(e1) corresponds to the linear fractional transformation z → 1/z. For the half-turn Rπ(e2), we have Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

260 18. Finite Mobius¨ Groups

λ 0 and µ 1, and the corresponding linear fractional transformation is z →−1/z. Finally, Rπ(N) corresponds to z →−z, the negative of the identity map, with λ i and µ 0. Adjoining the identity, we have

z →±z, ± 1 . z

This is the dihedral Mobius¨ group D2 of order 4. We obtain that D2 is a subgroup of the tetrahedral Mobius¨ group T. The four lines passing through the origin and the four vertices above intersect the opposite faces of the tetrahedron at the centroids. These are axes of symmetry rotations with angles 2π/3 and 4π/3. For√ example,√ for√ the rotation with angle 2π/3 and axis through (1/ 3, 1/ 3, 1/ 3), we obtain λ µ (1 + i)/2, so that the corresponding linear fractional transformation is ( + i)z − ( − i) z + i z → 1 1 . (1 + i)z + (1 − i) z − i In a similar vein, the 8 linear fractional transformations corresponding to these rotations are z + z − z + i z − i z →±i 1 , ±i 1 , ± , ± . z − 1 z + 1 z − i z + i These correspond to λ (±1 + i)/2,µ ±(1 + i)/2, and λ (±1 − i)/2,µ ±(1 − i)/2. Putting everything together, we arrive at the 12 elements of the tetrahedral M¨obius group T: z + z − z + i z − i z →±z, ± 1 , ±i 1 , ±i 1 , ± , ± . z z − 1 z + 1 z − i z + i We choose the octahedron in S2 such that its vertices are the six intersections of the coordinate axes with S2. This octahedron is homothetic to the intersection of the tetrahedron above and its reciprocal. As we concluded in Section 17, the octahedral group is generated by the tetrahedral group and a symmetry of the octahedron that interchanges the tetrahedron with its reciprocal. An example of the latter is the quarter-turn around the ﬁrst axis. This quarter-turn corresponds to the linear fractional transformation z → iz characterized by λ eπi/4 and µ 0. Thus, the 24 elements Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

18. Finite Mobius¨ Groups 261 of the octahedral Möbius group O are as follows: ik z + z − z → ikz, ,ik 1 ,ik 1 , z z − 1 z + 1 z + i z − i ik ,ik ,k , , , . z − i z + i 0 1 2 3 Finally, we work out the icosahedral Mobius¨ group I. We inscribe the icosahedron in S2 such that the North and South Poles become vertices. The icosahedron can now be considered as being made up of northern and southern pentagonal pyramids separated by a j pentagonal antiprism. The rotations S , j 0,...,4, S R2π/5(N), are symmetries of the icosahedron, and they correspond to the linear fractional transformations Sj : z → ωjz, j 0,...,4, where ω e2πi/5 is a primitive fifth root of unity. We still have the freedom to rotate the icosahedron around the vertical axis for a convenient position. We fine-tune the position of the icosahedron by agreeing that the second coordinate axis must go through the midpoint of one of the cross edges of the pentagonal antiprism. The half-turn U around this axis thus becomes a symmetry of the icosahedron, and as noted above, it corresponds to the linear fractional transformation U z →−1 . : z The rotations S and U do not generate the entire symmetry group of the icosahedron because they both leave the equator invariant. We choose for another generator the half-turn V whose axis is orthogonal to the axis of U and goes through the midpoint of an edge in the base of the upper pentagonal pyramid (Figure 18.1). Since U and V are (commuting) half-turns with orthogonal rotation axes, their composition W UV is also a half-turn whose rotation axis is orthogonal to those of U and V. With the identity, U, V, and W form a dihedral subgroup D2 of the icosahedral group I. To see what linear fractional transformations correspond to V and W, we need to work out the coordinates of the axis of V. To do this we first claim that the vertices of the icosahedron Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

262 18. Finite Mobius¨ Groups

V U

Figure 18.1

stereographically projected to Cˆ are

0, ∞,ωj(ω + ω4), ωj(ω2 + ω3), j 0,...,4.

Clearly, the poles correspond to 0 and ∞. The remaining ten vertices of the pentagonal antiprism projected to Cˆ appear in two groups of ﬁve points equidistantly and alternately distributed in two concentric circles (Figure 18.2). As in Section 17, we now think of the icosahedron as being the convex hull of three mutually orthogonal golden rectangles. Consider the golden rectangle that contains the North Pole as a vertex. The two sides of this golden rectangle emanating from the North Pole can be extended to Cˆ and give one projected vertex on each of the concentric circles. Considering similar triangles on the plane spanned by this golden rectangle, we see that the radii of the two concentric circles are τ and 1/τ (Figure 18.3). By Problem 5 (b) in Section 17, in terms of ω, these radii are 1 τ −ω2 − ω3 ω + ω4. and τ To ﬁnish the proof of our claim, we now note that the projected vertices have a 5-fold symmetry given by multiplication by ω.By construction, one of the vertices in the outer concentric circle must Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

18. Finite Mobius¨ Groups 263

Figure 18.2 be on the negative ﬁrst axis. The formula above for the projected vertices now follows by easy inspection. Returning to the main line, we now discuss how the projected vertices can be used to obtain coordinates of the axis of rotation for the half-turn V. Since the axis of U, the second coordinate axis, is orthogonal to the axis of V, the latter goes through the midpoint of −1 2 2 3 −1 3 2 3 the segment connnecting hN (ω (ω + ω )) and hN (ω (ω + ω )). −1 By the explicit form of hN (cf. Problem 1 (c) of Section 7), this

1/τ τ

Figure 18.3 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

264 18. Finite Mobius¨ Groups

midpoint is τ2 τ2 − 1 2 , 0, . τ2 + 1 τ2 + 1

Normalizing, we obtain that V Rπ(p0), where τ 1 p0 √ , 0, √ . τ2 + 1 τ2 + 1 Using the notation of Cayley’s theorem, we thus have i iτ λ √ and µ √ . τ2 + 1 τ2 + 1 We can rewrite λ and µ in terms of ω. A simple computation gives ω2 − ω3 λ − 1 √ ω − ω4 5 and ω2 + ω3 ω − ω4 µ √ ω − ω4 5 (cf. Problem 5 (c) of Section 17.)

Remark. The usual argument leading to λ and µ above involves computation of the angle of the axis of the rotation of V with a coordinate axis. In our approach we adopted Schli’s¨ philosopy and expressed all metric properties in terms of the golden section. The linear fractional transformation corresponding to the half- turn V ﬁnally can be written as (ω2 − ω3)z + (ω − ω4) V : z → . (ω − ω4)z − (ω2 − ω3) Composing this with U : z →−1/z, we obtain the linear fractional transformation corresponding to W UV: −(ω − ω4)z + (ω2 − ω3) W : z → . (ω2 − ω3)z + (ω − ω4) As noted above, W is also a half-turn, since the axes of U and V are orthogonal. Making all possible combinations of these linear Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

18. Finite Mobius¨ Groups 265 fractional transformations with those corresponding to multiplication by ωj, j 0,...,4, we ﬁnally arrive at the 60 elements of the icosahedral M¨obius group I: −(ω − ω4)ωkz + (ω2 − ω3) z → ωjz, − 1 ,ωj , ωjz (ω2 − ω3)ωkz + (ω − ω4)

(ω2 − ω3)ωkz + (ω − ω4) ωj ,j,k 0,...,4. (ω − ω4)ωkz − (ω2 − ω3) ♠ One final note. Under the aegis of Cayley’s theorem we constructed a list of finite Mobius¨ groups. Each of these groups G has a double cover G∗ in SU(2). G∗ is a finite group of special unitary matrices, and |G∗| 2|G|. G∗ is called the binary group associated to G. By definition, G∗ is the inverse image of G under the natural projection SU(2) → SU(2)/{±I}. Thus, we can talk about the ∗ ∗ binary dihedral group Dn, the binary tetrahedral group T , the binary octahedral group O∗, and the binary icosahedral group I∗. (We left the cyclic group out. Why?) We will return to these groups in Section 23 in a more geometric setting. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

. . . . . Detour in Topology: . . . . . Euler–Poincare´ . . . 19 . SECTION . Characteristic

♣ Euler’s theorem for convex polyhedra states that for any convex polyhedron, the alternating sum V − E + F is 2. This suggests that this quantity is a property that refers to something more general than the actual polyhedral structure. We thus venture away from convexity and try the alternating sum on nonconvex objects. We immediately run into difficulty, since we have not defined the concept of a nonconvex polyhedron; so far, all the polyhedra we’ve constructed have been convex. Fortunately, this is not a serious problem. We can simply say that a general polyhedron is the union of finitely many convex polyhedra, with the property that each two in the union are either disjoint or meet at a common face. This common face is deleted from the union. Figure 19.1 shows an example. Counting, we have V 16, E 32, and F 16, so V − E + F 0! (That Euler’s theorem fails for nonconvex polyhedra was first noticed by Lhuilier in 1812.) We begin to suspect that the fact that we ended up with 0 rather than 2 as before may have something to do with our new polyhedron having a “hole” in the middle. This is definitely a topological property! The alternating sum somehow detects the presence of “holes” (rather than the actual polyhedral structure), so that it is time to say farewell (but not goodbye!) to our Platonic solids. We do this 266 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

19. Detour in Topology 267

Figure 19.1 by circumscribing around each solid a sphere—a copy of S2—and projecting the faces radially to the sphere from the centroid. The faces projected to S2 become spherical polygons. We also realize that it does not matter whether we count V − E + F on the polyhedron or on S2. Turning the question around, we now start with S2. Consider a spherical graph on S2 with simply connected faces, and count V − E + F. Repeating the proof of Euler’s theorem, we realize that we always end up with 2. The conclusion is inevitable: This magic number 2 is a property of S2, not the actual spherical graphs! We say that the Euler–Poincare´ characteristic of S2 is 2, and write χ(S2) 2. One ﬁne point: The regular tetrahedron, octahedron, and icosahedron projected to S2 give spherical graphs whose faces are spherical triangles. When S2 is subdivided into spherical triangles, we say that S2 is triangulated. Going back to the neglected cube and dodecahedron, we see that (discarding regularity) we can split their faces into triangles without changing V −E+F, so that projecting these to S2 also give triangulations of S2. It is thus convenient to restrict ourselves to triangulations. We have now accumulated enough information to go beyond S2. Taking a look at Figure 19.1 again, we see that the appropriate surface to circumscribe about this polyhedron is the torus. The story is the same as before; the proof of Euler’s theorem leads to Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

268 19. Detour in Topology

the conclusion that the Euler–Poincare´ characteristic of the torus T2 is zero: χ(T2) 0. ♠ We now have great vistas ahead of us. We can take any general compact1 surface, triangulate it, and work out the Euler–Poincare´ characteristic. For example, looking at the “two-holed torus” depicted in Figure 19.2, we see that its Euler–Poincare´ characteristic is −2. In general, a “p-holed torus” or, more elegantly, a closed Riemann surface Mp of genus p, has Euler–Poincare´ characteristic

χ(Mp) 2 − 2p.

Another ﬁne point: You might be wondering why we did not consider the cylinder and all the noncompact examples H2/G, where G is Fuchsian, containing parabolic or hyperbolic isometries. The answer is that, at this point, we restrict ourselves to ﬁnite triangulations. Otherwise, the direct evaluation of V − E + F is impossible. Instead of going further along this line (which would lead us straight into homology theory, a branch of algebraic topology), we

Figure 19.2

1See “Topology” in Appendix C. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

19. Detour in Topology 269 explore some of the compact surfaces such as the Klein-bottle K2 and the real projective plane RP2. Recall that K2 is obtained from R2 by a discrete group generated by two glides with parallel axes. The fundamental domain again becomes a key player. In fact, if we triangulate the fundamental domain so that the side-pairing transformations map edges to edges, we obtain a triangulation of the quotient surface! In our case, we take the fundamental domain of the Klein bottle as described in Section 16 and triangulate it as in Figure 19.3. The simplicity of this is stunning; K2 can be obtained from two triangles by pasting their edges together appropriately! The Euler–Poincare´ characteristic is χ(K2) 1 − 3 + 2 0. Notice that the four vertices of the fundamental domain are iden- tiﬁed by the side-pairing transformations so that, on K2, we have only one vertex. Similarly, the two horizontal and the two vertical edges are identiﬁed, and the diagonal stays distinct. We do the same for RP2 S2/{±I} by triangulating S2 so that the antipodal map −I : S2 → S2 leaves the triangulation invariant; that is, it maps triangles to triangles, edges to edges, and vertices to vertices. The task is easier if we consider triangulation of the lower hemisphere only and make sure that the antipodal map restricted

Figure 19.3 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

270 19. Detour in Topology

Figure 19.4

to the equatorial circle maps edges to edges and vertices to vertices. A triangulation is given in Figure 19.4. We have V 3, E 6, and F 4. The Euler–Poincare´ characteristic is χ(RP2) 3 − 6 + 4 1. In Section 17, when studying symmetries of the octahedron, we came very close to a polyhedral model related to RP2. This model is called a heptahedron, since it has four triangular and three square faces. For the four triangles we take four nonadjacent faces of the octahedron. The four vertices complementary to a pair of opposite vertices of the octahedron are the vertices of a square. There are exactly three opposite pairs of vertices in the octahedron whose complements give the three square faces of the heptahedron (Figure 19.5). The three square faces intersect in three diagonals (connecting the three opposite pairs of vertices), and the diagonals intersect at the centroid of the octahedron in a triple point. Although the heptahedron is self-intersecting and “singular” at the vertices, it has Euler–Poincare´ characteristic of 1, since it has 6 vertices, 12 edges, and 7 faces. Furthermore, it is “nonorientable” in the sense that there is a triangle-square-triangle-square sequence of 4 adjacent faces that constitutes a polyhedral model of the Mobius¨ band. A natural question arises whether the heptahedron can be “deformed” into a smooth surface. Self-intersection cannot be elim- inated. In fact, it is not hard to prove that any compact surface that contains a Mobius¨ band cannot be realized as a smooth surface in R3 without self-intersections. As a ﬁrst attempt to make Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

19. Detour in Topology 271

Figure 19.5 a smooth heptahedral model, we consider the quartic (degree 4) surface given in coordinates p (a,b,c) ∈ R3 by the equation a2b2 + b2c2 + c2a2 abc

(see Figure 19.6). This is called the Roman surface, and it was studied by Steiner. Although this surface patterns the structure and the symmetries of the heptahedron, it still contains six singular points. (Where are they?) A somewhat better model can be obtained from the hemisphere model H of RP2 discussed at the end of Section 16. Indeed, identifying ﬁrst two pairs of equidistant antipodal pairs of points on the boundary circle of H and then pasting the remaining quarter circles together, we arrive at another model of RP2, algebraically given by the equation (a2 + 2b2)(a2 + b2 + c2) − 2c(a2 + b2) 0. In cylindrical coordinates a r cos θ, b r sin θ, and c, the equation reduces to r2 + (c − c(θ))2 c(θ)2,

Figure 19.6 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

272 19. Detour in Topology

Figure 19.7

where c(θ) 1/(1 + sin2(θ)). Thus the surface is swept by a rotating vertical circle with center at (0, 0, c(θ)) and radius c(θ) (see Figure 19.7). This surface self-intersects in the vertical segment connecting (0, 0, 1) and (0, 0, 2), and at the two endpoints it is still singular! The puzzling question whether RP2 has a realization in R3 as a smooth (self-intersecting) surface without singular points has been resolved by W. Boy,2 and the resulting surface is called the Boy’s surface. The basic building block of the Boy’s surface is a cylinder whose base curve is a leaf of the four-leaved rose given by the polar equation r sin(2θ),0≤ θ ≤ 2π. In Figure 19.8, the leaf is situated in the third quadrant of the plane spanned by the second and third axes in R3 so that the rulings of the cylinder are parallel to the first axis. It is important to observe that the cylinder has right “dihedral” angle along the first axis. We need a finite portion of the cylinder in the negative octant of R3 between the origin and the negative of the arc length of the leaf. We now rotate the cylinder around the axis R · (1, 1, 1) by 120◦ and 240◦. The configuration of the three cylinders is shown in Fig- ure 19.9 (a view from the positive octant) and in Figure 19.10 (a view from the negative octant). The cylinders intersect in three curves that meet in two triple intersection points. It is clear that this configuration can be made smooth along these curves, leaving

2D. Hilbert and S. Cohn-Vossen, Geometry and Imagination, Chelsea, New York, 1952, or W. Lietzmann, Visual Topology, Elsevier, 1969. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

19. Detour in Topology 273

Figure 19.8 the right dihedral angles intact. For the next step we bend the first cylinder such that the original ruling on the first axis is bent exactly to the shape of a leaf in the second quadrant of the plane spanned by the first and third axes (see Figure 19.11). Notice that the right dihedral angle can be retained, and that the hole created by the bending is congruent to the initial and terminal base leaves of the bent cylinder. We cover the hole by a flat leaf and perform the same bending and covering procedure to each of the three cylinders in the configuration in Figure 19.9. The final result is the Boy’s surface depicted in Figure 19.12. Since the dihedral angles were kept 90◦, this configuration is a

Figure 19.9 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

274 19. Detour in Topology

Figure 19.10

smooth self-intersecting surface with no singular points. The intersection is a closed curve consisting of three leaves forming a triple intersection point at the origin. A small strip cut from the surface along this curve gives a Mobius¨ band, and its complement is a topological disk. It follows that the Boy’s surface is a topological model of the real projective plane. It should be clear by now that the Euler–Poincare´ characteristic is a topological invariant; that is, two homeomorphic compact

Figure 19.11 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

19. Detour in Topology 275

Figure 19.12 surfaces have the same Euler–Poincare´ characteristic. This is because a triangulation on one surface can be carried over to the other by a homeomorphism. The big question is, of course, the converse: Given two compact surfaces with the same Euler–Poincare´ characteristic, are these surfaces homeomorphic? The answer is, in general, no. For example, the torus T2 and the Klein bottle K2 both have vanishing Euler–Poincare´ characteristic, but they are not homeomorphic, since the torus is orientable while the Klein bottle is not. It is a result of surface theory that two compact surfaces are homeomorphic iff their Euler–Poincare´ characteristics are equal and they are both orientable or nonorientable. This is indeed very beautiful, since a single number, plus the knowledge of orientability, characterizes the entire surface topologically! Many new examples of compact surfaces can be obtained by forming connected sums. Let M1 and M2 be compact surfaces. The connected sum M1 # M2 is obtained from M1 and M2 by cutting out open disks D1 ⊂ M1 and D2 ⊂ M2 and then pasting M1 − D1 and M2 − D2 together along the boundary circles ∂D1 and ∂D2.Tobe precise, by a disk here we mean the inverse image of a circular disk of R2 in the image of a chart. To show that we obtain a smooth surface requires some smooth- ing argument. It is also a technical matter (which we will not go into) that M1#M2 is unique up to homeomorphism; that is, its topological type does not depend on the disks chosen. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

276 19. Detour in Topology

When triangulations are given on both M1 and M2, then D1 and D2 can be chosen to be the interiors of some triangles. Forming M1#M2, we have the following changes for V1, E1, F1 (for M1) and V2, E2, F2 (for M2):

V V1 + V2 − 3 (since 3 pairs of vertices are identiﬁed);

E E1 + E2 − 3 (since 3 pairs of edges are identiﬁed);

F F1 + F2 − 2 (since 2 faces are deleted).

The Euler–Poincare´ charasteristic of the connected sum is therefore equal to

χ(M1 # M2) χ(M1) + χ(M2) − 2.

Before we investigate this formula, we note that M1#M2 is orientable iff both M1 and M2 are orientable. For our ﬁrst application of the formula above, we see that RP2 # RP2 is homeomorphic to the Klein bottle K2! Indeed, both are nonorientable, and χ(RP2 # RP2) 2χ(RP2) − 2 0asforK2. For a more direct argument, recall that RP2 is the Mobius¨ band and a disk pasted together along their boundaries. To form RP2 # RP2, we delete the corresponding two disks and paste the remaining Mobius¨ bands together along their boundaries. As a second example, we see that for any compact surface M the connected sum M # S2 is homeomorphic to M. As a third example, the connected sum of p copies of the torus T2 gives the p-holed torus, or a compact surface of genus p:

2 2 Mp T # ···# T (p times).

Here, equality means “homeomorphic.” Keeping this practice, we can write the ﬁrst two examples as

RP2 # RP2 K2 and M # S2 M.

(Algebraically, “twice” RP2 is K2, and S2 is the “zero” element for #. It may sound a little weird to add surfaces, but actually it is not as strange as it ﬁrst sounds.) Finally, we compile the following table: Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

19. Detour in Topology 277

Surface Euler–Poincare´ Characteristic T2 # ···# T2 (p times) 2 − 2p RP2 # ···# RP2 (p times) 2 − p RP2 #T2# ··· #T2 (p times) 1 − 2p K2#T2 # ··· #T2 (p times) − 2p

♠ Let f : M → N be a nonconstant analytic map between compact Riemann surfaces, and assume that M has genus p and N has genus q. By the table above, we have χ(M) 2 − 2p and χ(N) 2 − 2q. As shown in complex analyis, f : M → N is an n-fold covering with ﬁnitely many branch points in M. Near a branch point on M and near its f -image (called a branch value) local coordinates z and w can be introduced that vanish at the branch point (z 0) and at the branch value (w 0) such that in these coordinates, f has the form w zm. We call m − 1 the branch number of f at the branch point (cf. Section 15). The sum of all branch numbers is called the total branch number, and it is denoted by B. A standard argument3 shows that each point in N is assumed precisely n times on M by f , counting multiplicities. At a branch value, this means that n is equal to the sum of all m’s ([branch number plus 1]’s), where the sum is over those branch points that map to the given branch value. To relate the Euler–Poincare´ characteristics of M and N, triangulate ﬁrst N such that every branch value is a vertex of the triangulation. Let V, E, and F denote the number of vertices, edges, and faces of this triangulation. By Euler’s theorem, we have V − E + F χ(N) 2 − 2q. Now pull the triangulation on N back to a triangulation on M via f . Looking at what happens at a branch point reveals that the induced triangulation on M has − B vertices, nE edges, and nF faces. Once again by the Euler’s theorem, we have − B − nE + nF χ(M) 2 − 2p.

3See H.M. Farkas and I. Kra, Riemann surfaces, Springer, 1980. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

278 19. Detour in Topology

Comparing this with the previous formula, we obtain the Riemann– Hurwitz relation p n(q − 1) + 1 + B/2. There are a number of important consequences of this relation. For example, the total branching number B is always even; p 0 implies q 0; and if p, q ≥ 1 and p q, then f must have branch points. We will use the Riemann–Hurwitz relation for analytic maps f : S2 → S2, in which case it asserts that the total branching number is given by B 2n − 2.

Problems

1. Prove directly that T2 # RP2 is homeomorphic to RP2 # RP2 # RP2. 2. Check the computations leading to the table above.

Film

S. Levy, D. Maxwell and T. Munzner: Outside In, Geometry Center, University of Minnesota;AKPeters, Ltd. (289 Linden Street, Wellesley, MA 02181). Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

. . Detour in Graph . . . . . Theory: Euler, . . . . . Hamilton, and the . 20 . SECTION . Four Color Theorem

♣ If we lived in the early 1700s in the village of Konigsberg¨ (now Kaliningrad, formerly part of East Prussia), on nice sunny Sunday afternoons we would stroll along the river Pregel and walk over its seven bridges, which connect the banks and two islands in the river (Figure 20.1). We would overhear the people who pass by say that no one has ever been able to pass over all the bridges exactly once during one stroll. Some people would even say that this is impossible. (If you plan to visit the city, you may note that two more bridges have been built since then, one serving as a railway link. Our analysis however remains the same.) When this problem arrived at Euler’s desk around 1736, graph theory was born. Euler recognized that (as far as passing the bridges is concerned) it does not matter what our exact location is at any time during the walk as long as we know which of the two banks or two islands we are on. We can thus collapse these four pieces of land to four points (called vertices), and the bridges will become edges connecting these points. We arrive at the graph shown in Figure 20.2. The Konigsberg¨ bridge problem can thus be reformulated as follows: Does there exist a “walk” in this graph that passes along each edge exactly once? To be exact, we also have to specify whether 279 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

280 20. Detour in Graph Theory

Shopkeepers Blacksmith Wooden Bridge Bridge Bridge

Kneiphof Island Honey Bridge

Pregel

Green "Guts" Giblets Bridge Bridge

High Figure 20.1 Bridge

we wanted to arrive at the same spot we started at or not. In the former, we say that the walk is a “circuit”; in the latter, a “trail.” Due to the low number of vertices and edges (plus the axial symmetry), it is easy to see that there are no circuits or trails of this kind. Instead of doing a case-by-case check, we will follow Euler’s simple and powerful argument. Since this applies to any graph, it is now time to give some general deﬁnitions.

Figure 20.2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

20. Detour in Graph Theory 281

A graph G consists of a finite set V of vertices and a finite set E of edges. Each edge e ∈ E connects a two-element subset {v1,v2} from V. There may be multiple edges; that is, the same two vertices can be joined by more than one edge. Given a graph G,awalk in G is a sequence of vertices v1,v2,...,vk, where at least one edge e connects each pair of consecutive vertices {vi,vi+1}, i 1,...,k−1, in the sequence. In case of multiple edges, each two-element subset in a walk specifies a chosen edge. We usually denote a walk by v1 ···vk. We say that the walk joins the vertices v1 and vk. The graph is connected if there is a walk joining any two of its vertices. A walk is spanning if the vertices in the walk make up the whole of V. A walk is closed if v1 vk. If no vertices are repeated in a walk, we have a path.If no edges are repeated, the walk is a trail. A closed trail is called a circuit. Finally, a circuit that has at least one edge, and in which the only repeated vertex is v1 vk, is called a cycle. We see that the Konigsberg¨ bridge problem is equivalent to finding a circuit that includes each edge of the graph exactly once. Such circuits are called Eulerian. The graph is Eulerian if it contains an Eulerian circuit. Thus our problem is to decide whether the graph of the Konigsberg¨ bridge problem is Eulerian or not. There is a very simple criterion for the Eulerian property expressed in terms of the degree of vertices of the given graph. The degree of a vertex v in a graph G is the number of edges having v as an endpoint.

Theorem 12. A connected graph is Eulerian iff every vertex of the graph has even degree.

Proof. The proof is embarrassingly simple. Assume that G is Eulerian. Given an Eulerian circuit along which we are moving, every time we traverse a vertex v we do it by leaving an “entrance” edge and getting onto an “exit” edge. Since we have to traverse each edge exactly once, we discard these two edges, which contribute 2 in the degree. As we continue discarding, the degree of each vertex goes down by 2’s. Finally we run out of edges. Thus each vertex must have even degree. Conversely, assume that the degree condition Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

282 20. Detour in Graph Theory

holds and let us get started. At any point of our circuit-making trail, upon entering in a vertex (along an “entrance” edge), the question arises whether we have an exit edge that has not been used so far and can then be used to go on. The answer is yes, since each previous visit to v “consumed” exactly two edges, and if we were unable to go on the degree of v would be a sum of 2’s plus 1 corresponding to the entrance edge—an odd number. We are done.

It now takes only a second to realize that the graph in the Konigsberg¨ bridge problem has all odd-degree vertices, so it is not Eulerian! (This can be pushed a little further. We cannot traverse each bridge exactly once even if we drop the assumption that we arrive back at the same spot. For this, there must be exactly two odd-degree vertices (corresponding to departure and arrival), and all the other vertices must have even degrees.) What has this to do with Platonic solids? To explain this, we should consider vertices and edges as some sort of reciprocal no- tions. In an Eulerian circuit we must traverse each edge exactly once, but we can visit the vertices as many times as needed. Switch- ing the roles of edges and vertices, we now ask whether a given graph G has a spanning cycle; i.e., a circuit traversing each vertex of G exactly once. A spanning cycle is called Hamiltonian, and a graph possessing a Hamiltonian cycle is called Hamiltonian as well. Our first contact with the regular solids comes from Hamil- ton’s marketing ambitions (around 1857), which involved a wooden dodecahedron with each of its 20 vertices labeled with the name of a town. The puzzle was to find a circuit along the edges of the dodecahedron which passed through each town exactly once. A solution is given in Figure 20.3. We can say elegantly that the Schlegel diagram of a dodecahedron is Hamiltonian! Despite reciprocity of the Eulerian and Hamiltonian properties, there is no efficient algorithm to determine whether a graph is Hamiltonian. Instead of showing the subtleties of this problem, we switch to another problem closely related to regular solids. Recall that a graph was defined “abstractly” by its set of vertices and edges, and no reference was made to whether the graph can be realized (more elegantly, imbedded) in the plane such that vertices corre- Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

20. Detour in Graph Theory 283

Figure 20.3 spond to points and edges to continuous curves connecting these points. No edges are allowed to intersect away from the vertices. A graph with this property is called planar. Figure 20.4 shows two typical examples of nonplanar graphs. (In fact, a deep result of Kuratowski asserts that a graph is planar if it does not contain any subgraphs like these.)

Figure 20.4 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

284 20. Detour in Graph Theory

A representation of a connected planar graph on the plane is exactly what we used to prove Euler’s theorem on convex polyhedra. Thus, we immediately know that V − E + F 2. The structure of faces here depends on the speciﬁc representation while V and E do not. Toget a result for planar graphs independent of the planar representation, we therefore try to eliminate F from this formula. We arrive at the following:

Theorem 13. Let G be a connected planar graph with no multiple edges, and assume that E>1. Then E ≤ 3V − 6.

Proof. The case E 2 is trivial. We may therefore assume that E>2. Consider a speciﬁc representation of G in the plane. Since there are no multiple edges, each face is bounded by at least three edges. When counting edges this way, going from face to face, each edge (bounding two faces) is counted at most twice. We obtain that 3F ≤ 2E. Combining this with Euler’s theorem F 2 − V + E, we obtain 3F 6 − 3V + 3E ≤ 2E.

In 1852, a London law student named Francis Guthrie asked the following question: “Suppose you have a map, and want to color the various countries so that any two countries which share a common border always have different colors. What is the maximum number of colors that you might need?” Note that two different countries are allowed to have the same color provided that they do not share a common border. So, for example, Canada and Mexico may have the same color, but the United States must be colored differently from both Canada and Mexico. Also note that two countries may have the same color if they touch only at a corner, so if you have sixty-four countries arranged like the squares of a checker board, then the usual red and black coloring is legal. Guthrie conjectured that every map can be colored with at most four different colors. Guthrie was surely not the ﬁrst person to guess that four colors sufﬁce to color any map; some mapmaker must have made the Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

20. Detour in Graph Theory 285 same guess long before. But Guthrie seems to have been the first person to notice that this is a mathematical question. If it is really true that any map can be colored with four colors, then this ought to be a theorem with a mathematical proof. Guthrie himself was unable to find such a proof, but he was also unable to find a map that required five colors. So he told his brother Frederick about the problem. Frederick was studying mathematics at University Col- lege London with Augustus De Morgan (as Francis had done before deciding to study law) and he passed the problem on to De Mor- gan, who was also unable to find a proof or a counterexample. Over the years, the problem was passed around, mostly among British mathematicians, and in 1878 Arthur Cayley published the question in the Proceedings of the London Mathematical Society. The next year an amateur mathematician named Alfred Kempe published what he believed was a proof. Even Cayley was convinced by Kempe’s argument, but in 1890 Percy Heawood found a subtle flaw in Kempe’s reasoning. Heawood was, however, able to prove that no map requires more than five colors. No apparent progress was made for the next 86 years, and the four color problem became one of the most famous unsolved problems in mathematics. Finally, in 1976, Wolfgang Haken and Kenneth Appel, carrying out an idea suggested a few years earlier by Heesch, proved that Guthrie was right; every map can be colored with four colors. It is not surprising that such first-class mathematicians as De Morgan and Cayley were unable to solve this problem, because the proof that Haken and Appel discovered was far longer than any previous proof in the history of mathematics. The full proof, with all the details, has never been published, or even written down. If it were written down, it would certainly occupy millions, or perhaps billions, of pages. Naturally, Haken and Appel could not check all the details themselves; instead they wrote a computer program to check the proof for them. The program ran day and night for about two months! (Longer computer-assisted proofs have been discovered since 1976. For example, in 1990 Bren- dan McKay proved that at any dinner party with at least 27 guests, one can always find 3 people who all know each other or 8 people who are all strangers to each other, but that this is not always true with 26 guests. McKay’s proof consisted of checking every possible Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

286 20. Detour in Graph Theory

combination of acquaintance and non-acquaintance among the guests. Although he found an extremely efﬁcient way to check huge numbers of combinations simultaneously, he still had to run his computer for three years. McKay’s result is certainly a theorem, but somehow it strikes one as being of much narrower scope than the four color theorem. A handful of similar results exist in other branches of mathematics.) The problem of coloring maps is in fact a problem in graph theory, as we can see by assigning a vertex to each country and connecting two vertices with an edge whenever the corresponding countries share a common border. In this formulation, we must color the vertices of a graph, using different colors for any two vertices connected by an edge. The minimum number of colors needed to color a graph G in this way is called the chromatic number of G. If a map is drawn on a plane, then the corresponding graph is planar (see Problem 17). In other words, the four color theorem is equivalent to the statement that the chromatic number of a planar graph cannot be greater than four.

Remark. In Section 17, we described an algorithm to color the faces of the icosahedron with 5 colors such that the faces with the same color were disjoint. To obtain a 4-coloring we pick the four faces of a speciﬁc color group and recolor them using the remaining 4 colors such that no two faces with the same color meet at a common edge. (This can be done because each face has 3 adjacent faces and we have 4 colors.) This way we obtain a coloring of the faces of the icosahedron with 4 colors subject to the condition that faces with the same color can touch each other only at vertices. By reciprocity, this gives a coloring of the Schlegel diagram of the reciprocal dodecahedron with 4 colors in the sense discussed above.

The advantage of reformulating the four color problem as a question about planar graphs is that we can now apply Theorem 13. One slight obstacle remains: Theorem 13 requires that G have no multiple edges. But this is not a serious difﬁculty, because the chromatic number of a graph will not change if we remove all the multiple Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

20. Detour in Graph Theory 287 edges; all adjacent vertices will remain adjacent. (Suppose, for example, that five colors were required to color a map of Asia. The graph corresponding to this map does have multiple edges, because Russia and China share two common borders, one east and one west of Mongolia. If we were to eliminate the latter border by extending Mongolia to the west, then the map would still require five colors. In fact, conceivably it could require six, because Mon- golia would now share a border with Kazakhstan. So if there existed a counterexample to the four color theorem, we could surely find it among singly connected planar graphs.) An immediate consequence of Theorem 13 is the following:

Six Color Theorem. The chromatic number of a planar graph cannot be greater than 6.

Proof. First we must show that every planar graph with no multiple edges must include at least one vertex of degree 5 or less. (Before reading the rest of this proof, you may want to do Problem 8 at the end of this section.) Suppose there exists a planar graph G such that each vertex has degree 6 or more, with no multiple edges. If we add up the number of edges coming out of all the vertices, then each edge will be counted twice, because each edge connects two vertices. Therefore, 2E ≥ 6V and E ≥ 3V. This contradicts Theorem 13, which says that E ≤ 3V − 6. Now suppose there exist planar graphs with chromatic number greater than 6. Let G be such a graph with the minimum possible number of vertices (since V is positive, such a graph G must exist). Then if we remove a single vertex from G, along with its edges, the resulting graph G will have chromatic number 6 or less. In particular, we can remove one of the vertices v1 with five or fewer edges, and color the remaining vertices with six colors. Now restore the missing vertex v1. Every other vertex of G has already been colored; in particular the five (or fewer) vertices which share an edge with v1 have already been colored with five or fewer colors. Therefore, we can legally use the sixth color for v1, so contrary to our assumption, G can be colored with six colors. This contradiction implies Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

288 20. Detour in Graph Theory

that G does not exist, and every planar graph can be colored with six colors.

Heawood’s ﬁve color theorem1 and Haken and Appel’s four color theorem also follow from Theorem 13, but in a more roundabout way.

Five Color Theorem. Every planar graph has a chromatic number less than or equal to 5.

Proof. Suppose there exists a planar graph with chromatic number 6. Let G be the smallest such graph (that is, a graph with the smallest possible V). Clearly G cannot have any vertex with degree 4 or less, because we could then remove such a vertex, color the remaining (smaller) graph with five colors, and then replace the vertex, giving it a color different from that of its four neighbors. But, as we have seen, it follows from Theorem 13 that G must have a vertex of degree 5. Call one such vertex v1.Ifweremovev1, the other vertices of G can be colored with five colors. Furthermore, it must be the case that the five vertices adjacent to v1 will use all five colors, because if only four different colors were needed for those five vertices, then the fifth color could be used for v1. Call the five vertices v2, v3, v4, v5, and v6, moving clockwise in the plane around v1. This is illustrated in Figure 20.5; note that v2v3v4v5v6 is shown as a circuit with edges connecting the five vertices. There is no harm in making this assumption, because if the required edges were not originally present in G, we can surely add them without making G nonplanar or decreasing its chromatic number. Let us say that v2, v3, v4, v5, and v6 are colored red, yellow, green, blue, and violet respectively. Now suppose we were to remove all yellow, blue, and violet vertices from our graph, along with all the edges attached to those vertices. There are two possibilities: either v2 and v4 are still connected (that is, there is a path from v2 to v4 consisting only of red and green vertices), or they are not.

1This appeared in Volume 24 of the Quarterly Journal of Mathematics in 1890. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

20. Detour in Graph Theory 289

Figure 20.5

In the latter case, we can take the component of the graph connected to v4 and color all the green vertices red and the red vertices green, while leaving the component connected to v2 unchanged. We still have a legal coloring, because to get from one component to the other, you must pass through a region of yellow, blue, and violet vertices. But v4 is now red, and v2 is still red, so we can color v1 green. On the other hand, suppose that there is a path of red and green vertices connecting v2 and v4. In that case, there cannot be a path of yellow and blue vertices connecting v3 and v5, because the graph is planar, and the red-green path cannot cross the yellow-blue path. But if there is no such yellow-blue path, then we can swap the colors yellow and blue throughout the yellow-blue component connected Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

290 20. Detour in Graph Theory

to v5, while leaving the colors alone in the yellow-blue component connected to v3. We now have v3 and v5 both yellow, so we can color v1 blue. In either case, the entire graph, including v1, can be colored with ﬁve colors, contrary to our hypothesis. This contradiction establishes the theorem.

Kempe’s fallacious proof of the four color theorem was in two parts. He first proved that if G is the smallest graph (in terms of vertices) with chromatic number 5, then G has no vertices of degree 4 or less (see Problem 20). This part of the proof was correct. He then continued as follows: Let v1 be a vertex of G with degree 5 (by Theorem 13 and Problem 20, v1 must exist), and let v2, v3, v4, v5, and v6 be the vertices adjacent to v1, counting clockwise, as in the proof of the five color theorem. Since G is minimal, we can color all the vertices except v1 with four colors. Such a coloring must use all four colors for v2, v3, v4, v5, and v6; otherwise we could use the fourth color for v1. That means two of the five vertices must be the same color, and the other three different. The two vertices that are the same color cannot be adjacent, so we might as well assume that v3 and v6 are the same color (if not, just rename the vertices by moving the names around the circuit). In particular, let’s suppose that v3 and v6 are yellow, v2 is red, v4 is green, and v5 is blue (see Figure 20.6). Now suppose there is no red-green path connecting v2 and v4. Then we can recolor v4 red (also recoloring the red-green component connected to v4) while v2 remains red; v1 can then be green. If there is no red-blue path connecting v2 and v5, then we can recolor v5 red, and color v1 blue. The only other possibility is that a red-green path connects v2 to v4 and a red-blue path connects v2 to v5. In that case, there cannot be a yellow-blue path connecting v3 with v5, because, as one can see in Figure 20.6, such a yellow-blue path would have to cross the red-green path if it is to remain in the plane. Likewise, there cannot be a yellow-green path connecting v6 and v4, because such a path would have to cross the red-blue path. Therefore, we can swap the colors yellow and blue for all vertices inside the red-green path, and we can swap the colors yellow and green for all vertices inside the red-blue path. This changes v3 to blue (but leaves v5 blue) and changes v6 to green (but leaves v4 green). In that case, we can Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

20. Detour in Graph Theory 291

B G v 2 R R R v Y Y v 6 3

v 1 B B G G v v 5 4 R R Figure 20.6

color v1 yellow. So in all three cases, the entire graph, including v1, can be colored with four colors, contrary to our hypothesis. This contradiction establishes the theorem. Although Kempe’s proof doesn’t quite work, he was on the right track. He merely had to consider a few additional cases. Theorem 13 tells us that the average degree of the vertices in a planar graph is always less than 6. That means every planar graph has vertices of degree 5 (or less, but Problem 20 tells us that a minimal 5-color graph cannot have vertices of degree less than 5). But Theorem 13 actually tells us a good deal more than that. For example, it rules out the possibility that every vertex of degree 5 is surrounded by five vertices of degree ≥ 7, because in such a graph, the average vertex would have degree > 6. In other words, any minimal 5-color graph must include instances where (a) two vertices of degree 5 are adjacent or (b) a vertex of degree 5 is adjacent to a vertex of degree 6. The two configurations are shown in Figure 20.7, and they are said to make up an unavoidable set, which we define to be any set of subgraphs with the property that at least one of them is included in a minimal 5-color graph. Of course a single vertex of degree 5 is also, by itself, an unavoidable set, but it does us no good, because we cannot rule it out by using Kempe’s method of paths of alternating colors to show that a 5-color graph containing such a vertex cannot be minimal. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

292 20. Detour in Graph Theory

(a)

Figure 20.7 (b)

Suppose, however, it were possible to successfully carry out Kempe’s method on both the subgraphs shown in Figure 20.7. That is, suppose that for each of these subgraphs, we could prove that if a graph G containing that subgraph could be 4-colored except for one vertex in the subgraph, then it would always be possible to recolor G so that the missing vertex could also be legally colored. A subgraph on which we successfully carry out such a proof is said to be reducible. In that case, we would know that any graph containing subgraph (a) or (b) from Figure 20.7 could not be the minimal 5-color graph. But the minimal 5-color graph must include one of these subgraphs. Consequently, the minimal 5-color graph could not exist, and we would have proved the four color theorem. Unfortunately, Kempe’s method of paths of alternating colors cannot be successfully applied to the subgraphs in Figure 20.7. But it can be carried out with a number of slightly larger subgraphs, such as those in Figure 20.8. So perhaps we should try to ﬁnd a somewhat larger unavoidable set of larger subgraphs. All that we need to prove the four color theorem is an unavoidable set of reducible subgraphs. In 1969, Heesch put forward a statistical argument which suggested, but did not prove, that there exists an unavoidable set of about 8900 reducible subgraphs, each having no more than 18 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

20. Detour in Graph Theory 293

Figure 20.8 vertices around its perimeter. (In order to prove that a subgraph is reducible, you must consider every possible way to color its perimeter, so it is helpful to have the perimeter as short as possible.) He also developed methods, which could be carried out quite mechanically, for proving that a given set is unavoidable, and also for proving that a given subgraph is reducible. But Heesch had no way of ﬁnding this unavoidable set. And even if the set had somehow been handed to him (by divine revelation, say), it would have been completely impractical, given the computer hardware available in 1969, to prove that the set was unavoidable, or that each member of the set was reducible. Haken and Appel were able to sharpen Heesch’s statistical argument enough to convince themselves that an unavoidable set of reducible subgraphs could be constructed with no more than 14 vertices on the perimeter of each. This greatly decreased the Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

294 20. Detour in Graph Theory

time needed to prove reducibility. By 1976 computers were signif- icantly faster than in 1969, and so, after examining about 100,000 subgraphs, Haken and Appel managed to find an unavoidable set of 1936 reducible subgraphs, finally proving the four color theorem. Later they managed to decrease the size of their set to 1476. It is, to say the least, rather difficult to check such a proof, and no one seems to have actually done so. Even if one accepts the principle that a proof can be checked by computer, one would have to rewrite the software from scratch; simply rerunning the program that Haken and Appel wrote, even on a different computer, would mean little, since the program could contain errors. How- ever, in 1997 Neil Robertson, Daniel Sanders, Paul Seymour, and Robin Thomas were able to greatly simplify Haken and Appel’s proof by finding an unavoidable set of 633 reducible subgraphs. They also found a much faster algorithm for proving unavoidabil- ity, and they claim that in principle one could check the proof by hand in a few months (needless to say, no one has actually done this!), although a computer is still needed to check that each of the 633 subgraphs is in fact reducible. For details, including pictures of all 633 subgraphs, see Web Site 1 after the problems.

Problems

1. Call a graph G complete if any two distinct vertices in G are connected by a single edge. Given a complete graph G, show that G is Hamiltonian and 2E V(V − 1). 2. Show that if any vertex in a graph has at least degree 2, then the graph has a cycle. 3. Let G be a graph with no multiple edges. Show that if G has more than − 1)(V − 2)/2 edges, then G is connected. 4. Which Platonic solids have Hamiltonian Schlegel diagrams? 5. Deﬁne the Euler-Poincare´ characteristic of a graph as the number of vertices minus the number of edges. Show that if the graph is a tree (connected, with no cycle), then the Euler-Poincare´ characteristic is 1. (Note: The converse statement is also true.) 6. ♠ Show that the two typical examples of nonplanar graphs in Figure 20.4 cannot be imbedded into the plane. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 295

7. Use Theorem 13 to show that a complete planar graph can have at most four vertices. 8. Use Theorem 13 to show that a connected planar graph with no multiple edges must have at least one vertex of degree ≤ 5. 9. ♦ Prove Euler’s theorem for convex polyhedra using linear algebra. Assume that the connected graph G is directed, that is, the edges are ordered pairs of vertices of G. (Geometrically, this means that each edge has an arrow indicating its direction.) Index the vertices of G by 1, 2,... and the edges by 1, 2,...,E. Define the E × V edge-vertex matrix A of G as follows: If ek (vi,vj), i, j 1,... , k 1,...,E, the kth edge from the ith vertex to the jth vertex, then, in the kth row of A, define the ith entry to be −1, the jth entry +1, and zeros elsewhere. (a) Show that the kernel of A : RV → RE is one-dimensional and is spanned by (1, 1,...,1) ∈ RV . (b) Using (a), conclude that the rank of A is V − 1. (c) Assuming that G is planar, show that the number F − 1 of bounded faces is the dimension of the kernel of A+ : RE → RV , the transpose of A. (d) Using that A and A+ have the same rank, conclude that E − (F − 1) V − 1. 10. Complete the steps in von Staudt’s proof of Euler’s theorem for convex polyhedra. (a) Let G be a connected planar graph with V vertices, E edges, and F faces. Show that G contains a spanning tree T. (Spanning means that each vertex of G is a vertex of T.) (b) Define a graph T as follows: Pick a point from each region defined by G and call it a vertex of T. Thus T has F vertices. For each edge e of G not in T, choose a curve that avoids T and connects the two vertices of T that are contained in the regions meeting along e. Call this curve an edge of T. Prove that T isatree. (c) Use Problem 5 to count the edges of G by first counting those in T. Conclude that − 1) + (F − 1) E. 11. Show that a map of the United States cannot be colored with three different colors so that no two states with a common border have the same color. 12. Find a necessary condition for a map to be colorable with three colors. Is this condition also sufficient? 13. Suppose we require that two countries have different colors even if they only touch at one point. How many colors are needed to color a checker board? What is the maximum number of colors required to color any map? 14. As in Problem 13, we require that two countries have different colors even if they only touch at one point, but this time we specify that no more than four countries may come together at one point. Make a conjecture about the maximum number of colors required to color any such map. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

296 20. Detour in Graph Theory

15. Rephrase McKay’s result about the dinner party with 27 guests as a theorem about a complete graph on 27 vertices, where each edge is colored either red or green.

16. Show that the maximum number of colors required for any map drawn on the plane is the same as the maximum number of colors required for any map drawn on the sphere.

17. Prove that if a map is drawn on a plane, then the corresponding graph is planar.

18. What is the chromatic number of a complete graph on n vertices?

19. Give an example of a graph with chromatic number 4 which does not contain a complete subgraph of order 4. In other words, the four color theorem does not immediately follow from the fact that a complete planar graph cannot have ﬁve vertices.

20. Let G be the smallest graph (in terms of vertices) with chromatic number 5. Show that G has no vertices of degree 4 or less.

21. Find the ﬂaw in Kempe’s proof of the four color theorem. (This should take you much less than eleven years, even if you are not as smart as Cayley, because unlike Cayley you know that there is a ﬂaw.)

22. In this problem all graphs are imbedded in the real projective plane RP2. (a) Let G be a connected graph with no multiple edges imbedded in RP2 and assume that E>0. Using χ(RP2) 1 V − E + F, apply the proof of Theorem 13 to conclude that E ≤ 3V − 3. (b) Modifying the proof of the six color theorem, show that the chromatic number of a graph imbedded in RP2 cannot be greater than 6. (c) Show that the complete graph on six vertices can be imbedded in RP2. (Since the chromatic number of a complete graph on six vertices is 6 (see Problem 18), this shows that the upper bound in (b) is the best possible.)

23. Let M be a compact surface with Euler-Poincare´ characteristic χ(M) ≤ 0. Show that the chromatic number of any connected graph G imbedded in M cannot be greater than [(7 + 49 − 24χ(M))/2] using the following steps: Notice ﬁrst that the number inside the greatest integer function is a solution of the quadratic equation x2 − 7x + 6χ(M) 0. Rewrite this equation in the form 6(1 − χ(M)/x) x − 1 and apply the argument in the proof of Theorem 13 to conclude that there is a vertex of G with degree ≤ [x] − 1. Finally, modify the argument in the proof of the six color theorem for the present situation. (This result is due to Heawood. With the exception of the Klein bottle, the upper bound on the chromatic number is sharp.)2

2See G. Ringel and W.T. Youngs, “Solution to the Heawood map-colouring problem,” Proceedings of the National Academy of Sciences (U.S.A.), 1968. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Web Sites 297 Web Sites

1. www.math.gatech.edu/∼thomas/FC/fourcolor.html 2. www-groups.dcs.st-and.ac.uk/∼history/HistTopics/ Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Dimension Leap . . . 21 . SECTION .

♦ The success of developing complex calculus and the beauty of Riemann surfaces come about largely because we are able to multiply complex numbers and thus can form polynomials, power series, linear fractional transformations, etc. When we view complex numbers as planar vectors, complex multiplication becomes a speciﬁc operation F : R2 × R2 → R2. To extend our development to higher dimensions, we are now motivated to ﬁnd an operation F : Rn × Rn → Rn on the n-dimensional Euclidean real number space Rn. What conditions should F satisfy? Although opinions differ, most agree that F has to be bilinear, i.e., linear in both arguments:

F(a1p1 + a2p2,q) a1F(p1,q)+ a2F(p2, q),

F(p, a1q1 + a2q2) a1F(p, q1) + a2F(p, q2), n where p, p1,p2,q,q1,q2 ∈ R and a1,a2 ∈ R. (This corresponds to distributivity and homogeneity.) Instead of requiring that F be associative (which is hard to handle technically), we impose the condition that F be normed: |F(p, q)| |p|·|q|,p,q∈ Rn. 298 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

21. Dimension Leap 299

The advantage of this condition is clear. It connects algebra to geometry by simply declaring that the length of the product of two vectors must be the product of the lengths of the vectors! Our belief that this is the right condition is strengthened by our knowledge of complex multiplication, where this is a characteristic identity. A normed bilinear map F : Rn × Rn → Rn is called an orthogonal multiplication. Convinced as we are that the existence of an orthogonal multiplication is the key to developing higher dimensional analysis, our hopes are crushed by the following result of Hurwitz and Radon (c. 1898–1923).

Theorem 14. Orthogonal multiplications F:Rn × Rn → Rn exist only for n 1,2,4, and 8.

Remark 1. One is tempted to weaken the condition that F be normed as follows: A real division algebra structure on Rn is a bilinear map F : Rn × Rn → Rn such that F has no zero divisors, in the sense that p 0 q implies F(p, q) 0. This is absolutely necessary for our purposes; otherwise, we can’t form fractions. Theorem 14, however, can be generalized to the effect that real division algebra structure exists on Rn iff n 1, 2, 4, or 8.

Remark 2. You might object to all this by saying that we do have a multiplication that works in every dimension—the dot product! However, the dot product is not a genuine multiplication, because the dot product of two vectors is not a vector but a number. (Also, we cannot form triple products, and we have a lot of zero divisors.) If you are somewhat more sophisticated, you may ask why we don’t concentrate only on R3, where we have a multiplication × : R3 ×R3 → R3 3 given by the cross product of vectors. (Recall that, given v1,v2 ∈ R , the cross product v1 ×v2 is zero iff v1 and v2 are linearly dependent, and if they are linearly independent, then v1,v2,v1 × v2 (in this order) form a positively-oriented basis, with |v1 ×v2| equal to the area of the parallelogram spanned by v1 and v2.) The cross product is Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

300 21. Dimension Leap

1 not suitable either, since × is anticommutative (v1 ×v2 −v2 ×v1, 3 v1,v2 ∈ R )—in particular, we cannot even form squares!

A modern proof of Theorem 14 was given by Atiyah, Bott, and Shapiro. It relies on the classification of Clifford algebras and Clif- ford modules, a beautiful piece of modern algebra. It would not be too difficult to reproduce their work here, but since the proof uses the concept of the tensor product of algebras, we will go only as far as the definition of Clifford algebras. This is very much in the spirit of the Glimpses, and has the further advantage that the quaternionic identities will arise naturally. Let F : Rn × Rn → Rn be an orthogonal multiplication. Let e1 (1, 0,...,0), e2 (0, 1, 0,...,0) ,..., en (0,...,0, 1) denote the standard basis vectors in Rn. We define α n ui F(eα,ei) ∈ R ,i,α 1,...,n, using Greek and Latin indices to distinguish between first and second arguments. (F is not symmetric!) We claim that, for fixed α, {uα}n ⊂ n i i 1 R is an orthonormal basis. α First, ui is a unit vector, since α |ui | |F(eα,ei)| |eα|·|ei| 1. Second, let i k, i, k 1,...,n. On the one hand, we have α α 2 α 2 α 2 α α |ui + uk | |ui | +|uk | + 2ui · uk α α 2 + 2ui · uk . On the other hand, α α 2 2 |ui + uk | |F(eα,ei) + F(eα,ek)| 2 |F(eα,ei + ek)|

2 2 |eα| |ei + ek| 2. Combining these, we obtain α α ui · uk 0,i k,

1This does not mean that the cross product is not useful. In fact, it is the primary example of a Lie algebra structure on R3. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

21. Dimension Leap 301

i {uα}n ⊂ n which proves the claim. Similarly, for ﬁxed , i α 1 R is also an orthonormal basis. Next we ﬁx α, β and consider the orthonormal bases {uα}n {uβ}n i i 1 and k k 1 of Rn. Recall from linear algebra that the transfer matrix, denoted by Pβ,α, between these two orthonormal bases is an orthogonal Pβα (pβα)n matrix. In coordinates, ik i,k 1, and we have the change of bases formula n β βα α ui pik uk . k 1 Orthogonality of Pβα is expressed by Pβα(Pβα)+ I, where + stands for transpose. We now claim that, for α β, Pβα is skew-symmetric: (Pβα)+ −Pβα.

α α β β |ui + uk + ui + uk | α β α β 4 + 2ui · uk + 2uk · ui . We obtain the following fundamental identity: α β α β ui · uk + uk · ui 0,α β. This also holds for i k by orthogonality. Substituting the change of bases formula into this, we have n n α βα α α βα α ui · pkl ul + uk · pil ul 0, l 1 l 1 α and orthogonality of the ui ’s gives βα βα pki + pik 0. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

302 21. Dimension Leap

The claimed skew-symmetry follows. Combining skew-symmetry and orthogonality of Pβ,α, we obtain (Pβα)2 −I.

We see that Pβα is a complex structure on Rn. In general, a complex structure on Rn is a linear isometry J : Rn → Rn (represented by an orthogonal matrix) such that J 2 −I. Beyond the formal analogy with the complex identity i2 −1, a complex structure J can be thought of as a prescription for rotating vectors in Rn by π/2. Let us elaborate on this. First, given 0 v ∈ Rn, we claim that v and J(v) are orthogonal and of the same length. Indeed, |J(v)| |v| since J is an isometry, and v · J(v) J(v) · J 2(v) −J(v) · v −v · J(v);

so, v · J(v) must be zero. It is now clear how to rotate v by angle θ using the partial coordinate system {v, J(v)}; just deﬁne

Rθ(v) cos θ · v + sin θ · J(v), θ ∈ R.

Let us see if we can do this inductively. Let v1 v and denote by n σ1 the plane spanned by v and J(v). Let 0 v2 ∈ R be orthogonal to σ1. We have 2 J(v2) · v1 J (v2) · J(v1) −v2 · J(v1) 0 and

J(v2) · J(v1) v2 · v1 0,

so that J(v2) is also orthogonal to σ1. We obtain that the plane σ2 spanned by v2 and J(v2) is orthogonal to σ1. Continuing in this manner, we see that Rn can be decomposed into the sum of mutually orthogonal J-invariant planes: n R σ1 + σ2 +···+σm, and on each plane J acts by a quarter-turn. In particular, n 2m is even! All this can be put into a very elegant algebraic framework. Given a complex structure J on Rn,wecanmakeRn a complex vector Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

21. Dimension Leap 303 space by deﬁning multiplication of a vector v ∈ Rn to be given by a complex number z a + bi to be given by z · v a · v + b · J(v). We now return to our orthogonal multiplication F : Rn × Rn → Rn and investigate the fundamental identity above a little more. As before, we substitute the change of bases formula into the ﬁrst term of the fundamental identity, but now we switch α and β in the change of bases formula and substitute this into the second term. We obtain n n α βα α αβ β β ui · pkl ul + pkl ul · ui 0. l 1 l 1 α β Using orthogonality of the ui ’s and ui ’s again, we arrive at βα αβ pki + pki 0, or equivalently Pβα −Pαβ.

Finally, let α, β, and γ be distinct indices from 1,...,nand iterate the change of bases formula twice: n γ γβ β ui pik uk k 1 n n n γα α γα αβ β pil ul pil plk uk l 1 l 1 k 1 n n γα αβ β pil plk uk . k 1 l 1 Equating coefﬁcients, we obtain n γβ γα αβ pik pil plk . l 1 In matrix terminology, this means that Pγβ PγαPαβ, or, using skew-symmetry in the upper indices, PγαPβα −Pγβ. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

304 21. Dimension Leap

In particular, PγαPβα −PβαPγα,

so Pγα and Pβα anticommute! β,n Letting α n and introducing Jβ P , β 1,...,n− 1, we see that

J1,...,Jn−1 are pairwise anticommuting complex structures on Rn. Summarizing, we see that the existence of an orthogonal multiplication F : Rn × Rn → Rn implies that there exists a family n {J1,...,Jn−1} of anticommuting complex structures on R . ♥ These complex structures generate (under composition by multiplication) what is called a Clifford algebra. Since the elements of the Clifford algebra act on Rn as linear transformations, the vector space Rn becomes a Clifford module. Toprove Theorem 14 we would need to show that this is possible only for n 2, 4, and 8. Regretably, this is beyond the scope of these Glimpses. Let us mollify ourselves by instead taking a closer look at what happens in four dimensions.

Problems

2 1. Let J1 and J2 be complex structures on R . Show that either J1 J2 or J1 −J2. n 2. Let {J1,...,Jn−1} be a family of anticommuting complex structures on R . Prove that there exists an orthogonal multiplication F : Rn × Rn → Rn. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Quaternions . . . 22 . SECTION .

♦ We consider orthogonal multiplications for the case n 4, i.e., F : R4 × R4 → R4. We saw in the previous section that the existence of such F implies the existence of three complex structures 4 J1,J2,J3 on R that pairwise anticommute.

Proposition 4. Let {J1,J2,J3} be an anticommuting family of complex structures on R4. Then we have

J1 ◦ J2 ±J3.

Proof. Consider the linear isometry

U J1 ◦ J2 ◦ J3

4 of R . We claim that U commutes with each complex structure Jl, l 1, 2, 3, and U 2 I. Both claims follow by simple computations. For the ﬁrst, assuming l 1, we compute

U ◦ J J ◦ J ◦ J ◦ J J 2 ◦ J ◦ J J ◦ U. 1 1 2 3 1 1 2 3 1 305 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

306 22. Quaternions

For the second, we have 2 U (J1 ◦ J2 ◦ J3) ◦ (J1 ◦ J2 ◦ J3) J 2 ◦ J ◦ J ◦ J ◦ J 1 2 3 2 3 −J2 ◦ J 2 ◦ J 2 I. 1 2 3 Because U is an isometry whose square is the identity, it has only real eigenvalues, and they can only be ±1. Moreover, the eigenspaces

4 4 V+ {v ∈ R | U(v) v} and V− {v ∈ R | U(v) −v} are orthogonal, and together they span R4:

4 R V+ + V−.

By deﬁnition, U|V+ I and U|V− −I. That Jl commutes with U for each l 1, 2, 3, translates into Jl leaving V+ and V− invariant. 4 (In fact, if U(v) ±v, v ∈ R , then U(J1(v)) J1(U(v)) ±J1(v).) Thus, {J1,J2,J3} restricts to an anticommuting family of complex structures on V+ (and on V−). In particular, dim V+ is even, i.e., 0, 2, or 4. The middle dimension 2 cannot occur, since on a 2-dimensional vector space a complex structure is essentially determined by the orientation. Therefore, it is impossible for three complex structures to coexist with anticommutation (cf. Problem 1 4 of Section 21). Thus, V+ is either trivial or all of R . The same is true in reversed order for V−. Thus, U ±I, and we have J1 ◦J2 ◦J3 ±I. J J 2 −I Composing both sides by 3 from the right and using 3 ,we obtain J1 ◦ J2 ±J3 as claimed.

We have the liberty of choosing a sign for ±J3 without changing the entire structure (that is, the relations). We therefore assume that J1, J2 and J3 are arranged so that

J1 ◦ J2 J3. Summarizing, we obtained that the existence of an orthogonal multiplication F : R4 × R4 → R4 implies the existence of three linear 4 isometries J1,J2,J3 on R satisfying the relations: J2 J 2 J2 −I, 1 2 3 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

22. Quaternions 307 and

J1 ◦ J2 −J2 ◦ J1 J3,

J2 ◦ J3 −J3 ◦ J2 J1,

J3 ◦ J1 −J1 ◦ J3 J2, where the last two equalities can be derived from the ﬁrst. Now look at the following analogy: J 2 −I makes R2 a complex vector space with complex unit i (0, 1) satisfying i2 −1. Yielding to the obvious temptation, we introduce the vectors i (0, 1, 0, 0), j (0, 0, 1, 0), k (0, 0, 0, 1), in R4 and declare the rules for multiplication to be i2 j2 k2 −1 and ij −ji k, jk −kj i, ki −ik j.

After adding 1 (1, 0, 0, 0) to {i, j, k}, each vector q ∈ R4 can be written as a linear combination of 1,i,j,k: q a + bi + cj + dk, a, b, c, d, ∈ R, where we suppressed 1 from the notation. q expanded like this is called a quaternion. It is now clear how to multiply two quaternions using these identities. R4 equipped with this so-called quaternionic multiplication becomes a skew ﬁeld that we denote by H. (“Skew” means having noncommutative multiplication.) Multiplication of quaternions was introduced by Hamilton in 1843. According to the story, he had struggled with the problem of deﬁning multiplication of vectors in R3 since 1833, and his family took a great interest in this. Each morning at breakfast, his boys would ask, “Well, Papa, can you multiply triplets?” (meaning vectors in R3) and would receive the sad reply “No, I can only add and Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

308 22. Quaternions

subtract them.” Then, when strolling with his wife by Brougham Bridge in Dublin one day, it suddenly occurred to him that all the difficulties would disappear if he used quadruples—that is, vectors in R4. Overwhelmed with joy, he carved the identities above into the stonework of the bridge. Given a quaternion q a + bi + cj + dk ∈ H, in analogy with complex numbers it is customary to define the real part of q as !(q) a, and the pure part of q as P(q) bi + cj + dk. We also write q a + p with a ∈ R and p P(q). The conjugate of q a + p is then defined as q¯ a − p. Finally, we call a quaternion pure if its real part vanishes. The pure quaternions form the three- dimensional linear subspace H0 {q ∈ H | q¯ −q} of H spanned by i, j, k. Quaternionic multiplication satisfies the same identities as its complex brother. A word of caution is needed, however, for the identity

q1q2 q¯2q¯1,q1,q2 ∈ H,

in which the factors on the right-hand side get switched! Taking the analogy with complex arithmetic further, we now ask whether the ordinary dot product in R4 H can be written in terms of quaternions. Here it is:

q¯ q + q¯ q q · q 1 2 2 1 ,q,q ∈ H. 1 2 2 1 2

To show this, we ﬁrst note that the right-hand side is just !(q¯1q2). Setting q1 a1 + b1i + c1j + d1k and q2 a2 + b2i + c2j + d2k,we have

!(q¯1q2) !((a1 − b1i − c1j − d1k)(a2 + b2i + c2j + d2k))

a1a2 + b1b2 + c1c2 + d1d2,

since the mixed terms are all pure. The formula for the dot product follows. In particular, if q q1 q2, we obtain

|q|2 q · q qq,¯ Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

22. Quaternions 309 the usual Length2-Identity. With this, the quaternionic inverse of a nonzero quaternion q ∈ H can be written as q¯ q−1 . |q|2 This shows that H is indeed a skew ﬁeld. The 3-sphere S3 ⊂ H in quaternionic calculus is like the unit circle S1 ⊂ C in complex calculus. In fact, S3 {q ∈ H ||q| 1} constitutes a group under quaternionic multiplication; this is an immediate consequence of the fact that quaternionic multiplication is normed;

|q1q2| |q1|·|q2|,q1,q2 ∈ H. To check this, we compute 2 |q1q2| q1q2q1q2 q1q2q¯2q¯1 2 2 q1|q2| q¯1 q1q¯1|q2| 2 2 |q1| |q2| , 2 where we used the fact that reals (such as |q2| ) commute with all quaternions. What does S3 ⊂ H look like? The answer depends on whether you want an algebraic description or a “three-dimensional vision” in R4! We’ll use both approaches, beginning with the ﬁrst. Since we are already familiar with complex arithmetic, we just write q (a + bi) + j(c + di) a + bi + cj − dk z + jw,

z a + bi, w c + di ∈ C. We obtain that a quaternion is nothing but a pair of complex numbers; H C2. In terms of complex variables, quaternionic multiplication can be written as

q1q2 (z1 + jw1)(z2 + jw2)

z1z2 + jw1jw2 + z1jw2 + jw1z2

(z1z2 − w¯ 1w2) + j(w1z2 +¯z1w2), Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

310 22. Quaternions

2 so that, under the correspondence H C , multiplying q1 by q2 corresponds to matrix multiplication

z −w¯ z 1 1 2 . w1 z¯1 w2

Suppressing the indices, we say that multiplication by q z + jw corresponds to multiplication by the matrix

z −w¯ A . w z¯

Restricting q to S3 is equivalent to assuming |q|2 |z|2 +|w|2 1, and we see that A is special unitary, that is, A∗ A¯ + A−1

with determinant one. As observed in Section 18, these matrices constitute the important group SU(2) of special unitary 2 × 2 matrices. The correspondence that associates to the quaternion q ∈ S3 the special unitary matrix A is an isomorphism ϕ : S3 → SU(2) (it is clearly one-to-one and onto); that is, it satisﬁes

3 ϕ(q1q2) ϕ(q1) · ϕ(q2), q1,q2 ∈ S .

This follows by setting q1 z1 + jw1 and q2 z2 + jw2, and comparing the ﬁrst column of the product

z1 −w¯ 1 z2 −w¯ 2 ϕ(q1)ϕ(q2) w1 z¯1 w2 z¯2

with the complex expression of the product q1q2 above. The identiﬁcation SU(2) S3 allows us to use spherical concepts such as meridians of longitude and parallels of latitude on SU(2). Let

z −w¯ A(z, w) , |z|2 +|w|2 , w z¯ 1

be a typical element of SU(2), where we have displayed the dependence of A on z and w. The characteristic polynomial for A(z, w) Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

22. Quaternions 311 can be written as det(A(z, w) − tI) (z − t)(z¯ − t) + ww¯ t2 − (z +¯z)t + 1 t2 − 2!(z)t + 1. Since the real part !(z) of z is between −1 and +1, for ﬁxed r ∈ [−1, 1], we call {A(z, w) |!(z) r} the parallel of latitude at r. We see that two special unitary matrices are on the same parallel of latitude iff their characteristic polynomials are the same. The parallels of latitude corresponding to r 1 and r −1 are the single-point sets {I} and {−I}, which we may just as well call North and South Poles. For −1

eiθ 0 ∈ SU(2), θ ∈ R. 0 e−iθ

This meridian of longitude cuts the equator at

i 0 0 −i and its negative.

Remark. Let A ∈ SU(2). The conjugacy class of A is the set {BAB−1 | B ∈ SU(2)}. This spread is one line long. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

312 22. Quaternions

Since trace (BAB−1) trace A, it is clear that each conjugacy class is contained in a parallel of latitude. ♥ A somewhat more reﬁned analysis shows that the converse is also true, so that the parallels of latitude are exactly the conjugacy classes of matrices in SU(2).

The description of SU(2) S3 in terms of parallels of latitude (as conjugacy classes), though very pleasing, does not contain anything new about the geometry of S3. After all, the same geometric picture is valid in the one less dimension of S2! We now give a novel insight of the subtlety of the geometry of S3 absent in S2. ♦ In what follows, we parameterize S3 ⊂ C2 by two complex variables (z, w) satisfying |z|2 +|w|2 1. (Recall that z + jw is the quaternion corresponding to (z, w) ∈ C2.) Note that z runs on the ﬁrst and w on the second factor of C2 C × C. Consider the function f : S3 → R given by f(z, w) |z|2 −|w|2, (z, w) ∈ S3.

Since |z|2 +|w|2 1, we have −1 ≤ f ≤ 1. We now want to visualize the level sets 3 Cr {(z, w) ∈ S | f(z, w) r}, −1 ≤ r ≤ 1.

We have (z, w) ∈ Cr iff |z|2 −|w|2 r,

and, since |z|2 +|w|2 1, adding and subtracting yields 1 + r 1 − r |z|2 and |w|2 . 2 2

For r 1, we obtain that |z| 1 and w 0, so that C1 is the unit 2 circle in the ﬁrst factor of C C × C. Similarly, C−1 is the unit 2 circle in the second factor of C C × C; in particular, C1 and C−1 are perpendicular to each other. Now let −1

22. Quaternions 313 is the Cartesian product of two circles—a torus! We see that apart from the great circles C±1, the tori Cr, −1

1 iθ iθ S (z0,w0) {(e z0,e w0) | θ ∈ R} is the intersection of S3 with the 2-dimensional linear subspace in 2 C deﬁned by the equation zw0 − wz0 0. Since f(eiθz, eiθw) |eiθz|2 −|eiθw|2 |z|2 −|w|2 f(z, w), the second statement also follows. What is the quotient S3/S1? This should be two-dimensional since in S3 we are compressing great circles into points, so that the dimension must drop by one. We claim that S3/S1 and S2 are homeomorphic. Todo this, we need to understand how to associate 1 iθ iθ 2 2 to an orbit S (z0,w0) {(e z0,e w0) | θ ∈ R}, |z0| +|w0| 1, a unique point on the two-sphere S2. The easiest way to do this is to identify the projection map S3 → S3/S1. We introduce the Hopf map H : S3 → S2 given by H(z, w) (|z|2 −|w|2, 2zw)¯ ∈ R × C R3. This spread is one line long. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

314 22. Quaternions

First, note that H maps S3 to S2, since |H(z, w)|2 (|z|2 −|w|2)2 + 4|z|2|w|2 (|z|2 +|w|2)2 1 if (z, w) ∈ S3. Second, H is invariant under the action of S1, since

H(eiθz, eiθw) (|eiθz|2 −|eiθw|2, 2eiθzeiθw) (|z|2 −|w|2, 2zw)¯ H(z, w). Thus, H maps each orbit of S1 in S3 into a single point. To show that S3/S1 S2, we need to prove that the orbits are precisely the inverse images of points from S2. In other words, we have to show that whenever H(z1,w1) H(z2,w2), the points (z1,w1) and 1 (z2,w2) are on the same orbit under S . Now the fact that the Hopf images are equal translates into

2 2 2 2 |z1| −|w1| |z2| −|w2| and

z1w¯ 1 z2w¯ 2.

The ﬁrst equality means that (z1,w1) and (z2,w2) are on the same Clifford torus, say Cr,sowehave 1 + r 1 − r |z |2 |z |2 and |w |2 |w |2 . 1 2 2 1 2 2 iθ iϕ Letting z2 e z1 and w2 e w1, we substitute these back to the second equation and obtain i(θ−ϕ) z1w¯ 1 e z1w¯ 1,

and so (exluding the trivial cases when z1 0orw1 0, which can be handled separately) ei(θ−ϕ) 1 follows. By the periodicity property of the exponential function, θ iθ and ϕ differ by an integer multiple of 2π. Thus w2 e w1, and we are done. ♠ We can deﬁne the complex projective n-space CPn using the same construction as for the real projective n-space RPn (Problem 5 in Section 16). CPn comes equipped with the natural projection Cn+1 −{0}→CPn associating to a nonzero complex vector p ∈ Cn+1 This spread is one line long. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 315 the complex line C · p minus the origin. Restricting the projection to the unit sphere S2n+1 ⊂ Cn+1 ( R2n+2), we obtain a map S2n+1 → CPn that associates to p ∈ S2n+1 the set of multiples eiθp, θ ∈ R, that span the complex projective point corresponding to p. This is better understood when we define a natural action of S1 ⊂ C on S2n+1 ⊂ Cn+1 given by multiplying complex vectors by eiθ ∈ S1. Then S2n+1 → CPn is nothing but the orbit map. Note that each orbit of S1 on S2n+1 is a great circle. In particular, S2n+1 can be thought of as composed of circles attached to every point of CPn! For n 1, the orbit map S3 → CP1 is nothing but the Hopf map. Hence CP1 can be identified with S2! This is not too surprising, though; the ordinary points [z : w] ∈ CP1 with complex homogeneous coordinates z, w ∈ C are exactly those with nonzero w. Thereby they can be made to correspond to ratios z/w ∈ C, and the only ideal point is [1 : 0], corresponding to ∞. We thus have CP1 C ∪{∞} S2! We close this section with an advanced remark. If we look at two orbits of the action of S1 on S3 (on the same Clifford torus), it is apparent that they are “linked” in S3. This means that the Hopf map H : S3 → S2 cannot be deformed continuously through maps into a constant map S3 → S2 that sends the whole S3 to a single point. We express this by saying that H is homotopically nontriv- 2 ial, or, even more formally, that the third homotopy group π3(S ) is nontrivial. This was a pioneering result of Hopf’s during the early development of homotopy theory, since because of the apparent similarity between homotopy and homology theories, one 2 expected to find π3(S ) to be trivial since the third homology group 2 H3(S ) 0.

Problems

1. Consider the 3-dimensional subspace V C × R · k ⊂ H. Show that the map q → kqk−1, q ∈ V, leaves V invariant. Show that this map is an opposite isometry of V and describe it geometrically using V R3.

2. Let q1,q2 ∈ H0 be purely imaginary quaternions. Show that the quaternionic product q1q2 ∈ H projected to R · 1 is the negative of the dot product of q1 3 and q2 considered as spatial vectors under the identiﬁcation H0 R . Show that q1q2 projected to H0 is the cross product of q1 and q2. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

316 22. Quaternions

2 2 2 3. Use the quaternionic identity |q1| |q2| |q1q2| , q1,q2 ∈ H, to prove the four square formula:1 (a2 + b2 + c2 + d2)(a2 + b2 + c2 + d2) 1 1 1 1 2 2 2 2 2 (a1a2 − b1b2 − c1c2 − d1d2) 2 + (a1b2 + b1a2 + c1d2 − d1c2) 2 + (a1c2 − b1d2 + c1a2 + d1b2) 2 + (a1d2 + b1c2 − c1b2 + d1a2) . 4. Use the four square formula of Problem 3 to show that if a and b are both sums of four squares of integers, then ab is also a sum of four squares of integers. (Lagrange proved in 1772 that any positive integer can be expressed as the sum of four squares of integers (cf. Problem 7 of Section 5). This problem has a rich and complex history. In 1638, Fermat asserted that every positive number is a sum of at most three triangular numbers, four squares, five pentagonal numbers, and so on (cf. Problem 11 of Section 2). In 1796, in one of the earliest entries in his mathematical diary, Gauss recorded that he had found a proof for the triangular case. The general problem was resolved by Cauchy in 1813.) 5. Define the quaternionic Hop map H : H2 → R × H by H(p, q) (|p|2 −|q|2, 2pq),¯ p, q ∈ H, and show that H maps the unit sphere S7 ⊂ H2( R8) onto the unit sphere S4 ⊂ R × H( R5). Verify that the inverse image of a point in S4 under H : S7 → S4 is a great 3-sphere S3 in S7. 2 6. Associate to the spatial rotation with axis R · p0,p0 (a0,b0,c0) ∈ S , and angle θ ∈ R, the antipodal pair ±(a + bi + cj + dk) ∈ S3 of unit quaternions, where a cos(θ/2), b c0 sin(θ/2), c b0 sin(θ/2), and d a0 sin(θ/2). (a) Describe the set of spherical rotations that correspond to parallels of latitude in S3. (b) ♠ Study the correspondence between the group of spherical rotations, the quotient group S3/{±1}, and Möb+(Cˆ ).

Web Sites

1. www.geom.umn.edu/∼banchoff/script/b3d/hypertorus.html 2. www.maths.tcd.ie/pub/HistMath/People/Hamilton/Letters /BroomeBridge.html

1This identity appears in a letter Euler wrote to Goldbach in 1705. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... 3 . . Back to R ! . . . 23 . SECTION .

♦ After our frustration over the nonexistence of orthogonal multiplications in three dimensions, we now try to incorporate R3 into our skew ﬁeld H. The symmetric role of the quaternionic units i, j, and k indicates that R3 should sit in H as the linear space H0 {q ∈ H | q¯ −q} of pure quaternions. As we did in our in- vestigations of complex arithmetic, we now want to see what kind of geometric transformations arise in R3 from quaternionic multiplication restricted to H0. As an elementary example, conjugation in H restricts to the antipodal map in H0!

Theorem 15. q ∈ q →±q qq−1 q ∈ Let 0 0 H. Then the transformations 0 0 , H, are linear isometries that leave H0 invariant.

q ∈ q →−q qq−1 (1) If 0 H0, then the restriction of 0 0 to H0 is reﬂection to the plane orthogonal to q0.

(2) If q0 a0 + p0,0 a0 ∈ R,0 p0 ∈ H0, then the restriction of q → q qq−1 · p <θ<π 0 0 to H0 is rotation with axis R 0 and angle 0 given by tan (θ/2) |p0|/a0. 317 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

318 23. Back to R3!

Proof. |q qq−1| |q ||q||q |−1 |q| Since 0 0 0 0 , it is clear that the transforma- q →±q qq−1 4 tions 0 0 are linear isometries of H R . The invariance of H0 under these transformations can be proved directly. Instead, we will develop a criterion for a quaternion to belong to H0, which will turn out to be useful later on. 2 We claim that q ∈ H0 iff q is a nonpositive real number. Indeed, given q a + p, a ∈ R,0 p ∈ H0 (we may assume that p is nonzero, since otherwise the claim follows), we have q2 a2 + 2ap + p2. The last term p2 is a nonpositive real number, since p2 (bi + cj + dk)2 −(b2 + c2 + d2)

(the mixed terms cancel because of anticommutativity of i, j, k). Thus, the pure part of q2 is equal to 2aP(p) 2ap, and this is zero iff a 0. Returning to the proof of invariance of H0, we need to show that q ∈ q qq−1 ∈ H0 implies 0 0 H0. Using the criterion just proved, this is equivalent to the statement that whenever q2 is a nonpositive real (q qq−1)2 number, then so is 0 0 . We compute the latter as (q qq−1)2 q qq−1q qq−1 q q2q−1 0 0 0 0 0 0 0 0 q2q q−1 q2, 0 0 where the last but one equality is because a real number commutes with all quaternions. Invariance of H0 follows. The transformations q →±q qq−1 p ±p 0 0 send 0 to 0, since q p q−1 q (q − a )q−1 q q q−1 − a q q−1 q − a p . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Since these are also isometries, the plane P perpendicular to p0 remains invariant. Let q ∈ P and compute the cosine of the angle θ q q qq−1 between and 0 0 (Figure 23.1). Using the dot product formula, we have q · (q qq−1) 1 cos θ 0 0 (q(q¯ qq−1) + q qq−1q) |q|2 2|q|2 0 0 0 0

1 (qq¯ qq¯ + q qq¯ q) 2 2 0 0 0 0 2|q| |q0| Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

23. Back to R3! 319

p 0

θ − q q qq 1 0 0

Figure 23.1

1 (qq¯ qq¯ + q q¯q¯ q) 2 2 0 0 0 0 2|q| |q0| − 1 (qq qq¯ + q qq¯ q), 2 2 0 0 0 0 2|q| |q0| where we used q¯ −q. To rewrite this into a more convenient form, we now use orthogonality of q and p0. We claim that orthogonality implies the commutation relation

qq0 q¯0q.

Indeed, since q and p0 are orthogonal, q · p0 (qp¯ 0 + p¯ 0q)/2 −(qp0 + p0q)/2 0, so q and p0 anticommute. With this, we have

qq0 q(a0 + p0) qa0 + qp0 a0q − p0q (a0 − p0)q q¯0q. Returning to the main computation, 1 cos θ − (qq qq¯ + q qq¯ q) 2 2 0 0 0 0 2|q| |q0| 1 − (q¯ q2q¯ + q q2q )(q2 is real!) 2 2 0 0 0 0 2|q| |q0| q2 − (q¯ 2 + q2) 2 2 0 0 2|q| |q0| 1 a2 + p2 !(q2) 0 0 . |q |2 0 a2 − p2 0 0 0

For case (1) of the theorem, we have a0 0 (and p0 q0), so θ − θ π q → q qq−1 that cos 1 follows. Thus and 0 0 restricted to Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

320 23. Back to R3!

P q →−q qq−1 is a half-turn. Incorporating the negative sign, 0 0 is identity on P and, as we have seen earlier, it sends q0 to −q0. Thus q →−q qq−1 P 0 0 is rection in the plane . p q → q qq−1 For case (2) of the theorem, 0 is left ﬁxed, and 0 0 restricted to P is rotation with angle 0 <θ<πwhere a2 + p2 cos θ 0 0 a2 − p2 0 0 Now the trigonometric identity tan2(θ/2) (1 − cos θ)/(1 + cos θ) gives tan(θ/2) |p0|/a0. The theorem follows.

Theorem 15 means that spatial rections and rotations can be 3 obtained from quaternionic multiplications restricted to R H0. These make up the group of all linear isometries of R3 (leaving the origin ﬁxed), that is, they make up the orthogonal group O(R3). Returning to our original aim, we now place the Platonic solids in H0 (with centroids at the origin) and express the symmetries in terms of quaternions. In this way, we obtain a very transparent description of the symmetry groups of Platonic solids. Before we do this, let us formalize what we just said about representing the elements of O(R3) by quaternions. For simplicity, we restrict ourselves to direct linear isometries that constitute the special orthogonal group SO(R3), a subgroup of O(R3).

Theorem 16. 3 The map ψ that associates to each unit quaternion q0 ∈ S the q → q qq−1 transformation 0 0 restricted to H0 is a surjective group homomorphism ψ : S3 → SO(R3) with kernel ker ψ {±1}.

Proof. By Theorem 15, if q0 ±1, then q0 deﬁnes a rotation, an element 3 of SO(R ). On the other hand, q0 ±1 deﬁnes the identity element in SO(R3) so that ψ maps into SO(R3). It is clear that ψ is Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

23. Back to R3! 321 a homomorphism of groups, and, by what we just said, ±1arein the kernel of ψ. By Theorem 10 of Section 16 and Theorem 15, ψ is onto since all elements in SO(R3) are rotations. It remains to show that the kernel of ψ is exactly {±1}. Let q0 ∈ ker ψ, that q qq−1 q q ∈ q is, 0 0 for all H0. Equivalently, 0 commutes with all pure quaternions. Writing this condition out in terms of i, j, and k, 3 we obtain that q0 must be real. Since it is in S , it must be one of ±1.

Remark. ♥ Theorem 16 implies that the group S3 of unit quaternions modulo the normal subgroup {±1} is isomorphic with the group SO(R3) of direct spatial linear isometries. The quotient group S3/{±1} is, by deﬁnition, the group of right- (or left-) cosets of {±1}. A right-coset containing q ∈ S3 thus has the form {±1}q {±q}. Thus, topologically, S3/{±1} can be considered as a model for the projective space RP3. The classical model of RP3 is the same as the model of RP2 discussed in Section 16; that is, a projective point is a line through the origin of R4, etc. By Theorem 16 above, RP3 can be identiﬁed by the group of direct spatial isometries SO(R3)!

♦ Let us explore some concrete settings. As we learned in Section 22, a quaternion q can be represented by a pair of complex numbers (z, w) ∈ C2 via q z + jw. The second variable w ∈ C in this representation corresponds to jw ∈ jC, and we see that jC is the complex plane in H0 spanned by the vectors j and k( ij). The complex unit i ∈ H0 is orthogonal to jC in H0. As an example, we now describe a rotation in H0 with axis R · i and angle θ,0<θ<π. According to Theorem 15, this rotation is described by the quaternion q0 a + p0, a ∈ R,p0 ∈ H0, via q → q qq−1 q ∈ p i q 0 0 , H0. Since 0 , 0 happens to be a complex number. The condition on the angle can be written as tan(θ/2) |i|/|a| 1/|a|, so that, choosing a positive, we have q0 cot(θ/2) + i.We normalize q0 to a unit: q cot(θ/2) + i θ θ 0 cos + i sin . 2 |q0| cot (θ/2) + 1 2 2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

322 23. Back to R3!

We obtain that rotation with axis R · i and angle θ corresponds to the complex number eiθ/2 viewed as a quaternion. To check that this is correct, we compute q iq−1 eiθ/2ie−iθ/2 i, 0 0 so that i is kept ﬁxed. Moreover, eiθ/2 acts on jw ∈ jC as eiθ/2jwe−iθ/2 eiθjw

and this is indeed rotation by angle θ on jC! Notice now that eiθ/2 and its negative −eiθ/2 ei(π+θ/2) represent the same rotation, since by Theorem 16, there is a two-to-one correspondence between unit 3 quaternions in S and linear isometries acting on H0. We now consider the cone Cn in H0 with vertex hi (where h>0is the height) and base jPn, where the regular n-sided polygon Pn ⊂ C is placed in jC (Figure 23.2). lπi/n We see that the cyclic group C2n {e | l 0,...,2n − 1} of unit quaternions is a double cover of the symmetry group + lπi/n Symm (Cn), in the sense that, for l 0,...,n − 1, e and lπi/n (n+l)πi/n −e e correspond to the same rotation in H0. The same argument can be used for the prism Pn given by h h − , i × jPn. 2 2

Besides the elements of C2n, we have half-turns at the vertices and the midpoints of edges of jPn. They correspond to the quaternions {jeliπ/n | l 0,...,2n − 1}.

jPn

Figure 23.2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

23. Back to R3! 323

Putting these elements together, we obtain the so-called binary dihedral group

∗ lπi/n lπi/n Dn {e | l 0,...,2n − 1}∪{je | l 0,...,2n − 1} + and this is a double cover of Symm (Pn). We will now turn to the Platonic solids and derive double covers of the tetrahedral, octahedral, and icosahedral groups in terms of quaternions. We begin with the tetrahedral group. We position the regular 3 tetrahedron T in R H0 as in Section 17 with vertices i + j + k, i − j − k, −i + j − k, −i − j + k.

Symm+(T ) is generated by the three half-turns around the coordinate axes and by the rotation around i + j + k by angle 2π/3. The ﬁrst three correspond to the three pairs of quaternions ±i, ±j, ±k, and the latter is given by the quaternion (normalized to belong to S3) 1 ± √ (a + i + j + k), a > 0, a2 + 3 where √ |i + j + k| (π/ ) 3 . tan 3 a a We obtain that a 1, so the pair of quaternions corresponding to rotation at the front vertex i + j + k is ( + i + j + k) ± 1 . 2 Putting these together, we see that the binary tetrahedral group deﬁned by ( ) (±1 ± i ± j ± k) T∗ {±1, ±i, ±j, ±k}∪ ⊂ S3 2 is a double cover of Symm+(T ). Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

324 23. Back to R3!

∗ 3 Let us visualize T ⊂ S through the Clifford* decomposition of S3 S3 C discussed in Section 21. Recall that −1≤r≤1 r, where ( ) + r − r 2 2 1 2 1 Cr (z, w) ∈ C |z| , |w| . 2 2

3 4 C±1 are orthogonal great circles cut out from S ⊂ R by the coordinate planes spanned by the ﬁrst two and the last two axes. For 3 −1

so that they are all in the middle Clifford torus C0. We now view C0 as [0, 2π]2 with opposite sides identiﬁed as in Section 15. In this representation (z, w) ∈ C0 corresponds to the point (arg(z), arg(w)) ∈ [0, 2π]2. (Here we take the value of arg in [0, 2π].) Working out the arguments of (±1 ± i)/2, we obtain all odd multiples of π/4 on [0, 2π]. Thus, on C0, the binary tetrahedral group has sixteen points whose coordinates are odd multiples of π/4. Putting these together, we get 4 + 4 + 16 24 |T∗| points! (See Figure 23.3.) The other examples are treated similarly. The octahedron O is 3 placed in R H0 with vertices ±i, ±j, ±k,

Figure 23.3 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

23. Back to R3! 325 and the rotations around each vertex with angle π/2 are given by the quaternions ± ± i ± ± j ± ± k √1 , √1 , 1√ . 2 2 2 Since Symm+(T ) ⊂ Symm+(O) (recall that O can be obtained from T by four truncations), these and T∗ generate the binary octahedral group 1 + i O∗ T∗ ∪ √ T∗ ⊂ S3. 2 √ This is a double cover of Symm+(O). Notice that (1 + i)/ 2 eiπ/4 corresponds to rotation around the imaginary axis R · i with angle π/2 and this is precisely the isometry that carries T into its reciprocal T ◦. We thus see the algebraic counterpart of the geometric argument used to determine the octahedral group in Section 17. Multiplication by eπi/4 has the effect of adding π/4 to the parameters arg (z) and arg (w) of the Clifford decomposition of S3. Note D∗ ∗ that 4 becomes a subgroup of O . The entire binary octahedral group is depicted in Figure 23.4. Finally, working out all quaternions that give icosahedral rotations, we arrive at the double cover of Symm+(I): I∗ T∗ ∪ σT∗ ∪ σ2T∗ ∪ σ3T∗ ∪ σ4T∗, σ 1 (τ+i+j/τ) τ where 2 and is the golden section (see Section 17). This is called the binary icosahedral group. We omit the somewhat gory details (cf. Problem 2). Note again the perfect analogy between this algebraic splitting of I∗ and the geometric description of the icosahedral group in terms of the ﬁve circumscribed tetrahedra in Section 17. To place I∗ in the Clifford setting needs a bit of computation. It is clear that the elements ±ωk and ±jωk, k 0,...,4, make up

Figure 23.4 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

326 23. Back to R3!

the vertices of two copies of a regular decagon, one inscribed in ∗ C−1, the other in C1. These account for 20 elements of I . For the remaining 100 elements, we have 1 |z|2 −|w|2 ±√ (|ω − ω4|2 −|ω2 − ω3|2) 5 1 ∓√ ((ω − ω4)2 − (ω2 − ω3)2) 5 1 ±√ (ω + ω4 − ω2 − ω3) 5 ±√1 1 + τ ±√1 . 5 τ 5 We see that these elements (in two groups of 50) are on the two C √ Clifford tori ±1/ 5. Calculating the arguments is now easy, since ω − ω4 ω2 − ω3 C √ and are both purely imaginary. On 1/ 5,we obtain π kπ π lπ π kπ π lπ 3 + 2 , + 2 , + 2 , 3 + 2 , 2 5 2 5 2 5 2 5 k, l C √ where are integers modulo 5. Similarly, on −1/ 5 we have π kπ π lπ π kπ π lπ + 2 , + 2 , 3 + 2 , 3 + 2 , 2 5 2 5 2 5 2 5 where again k, l are integers modulo 5. The entire binary icosahedral group I∗ is depicted in Figure 23.5. Thus, quaternions can be used to describe the symmetry groups of the Platonic solids in a simple and elegant manner. Any computation involving these groups can be carried out using quaternionic arithmetic.

Figure 23.5 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

23. Back to R3! 327

♥ One ﬁnal algebraic note: In searching for quaternionic representation of the symmetries of the Platonic solids, we found the following ﬁnite subgroups of S3:

2πli/n 1. Cn {e | l 0,...,n− 1}; D∗ C ∪ jC D∗ {± , ±i, ±j, ±k} 2. n 2n 2n; in particular, 2 1 ; ∗ D∗ ∪{(± ± i ± j ± k)/ } 3. T 2 1 √ 2 ; 4. O∗ T∗ ∪ ((1 + i)/ 2)T∗; ∗ ∗ ∪ σ ∗ ∪ σ2 ∗ ∪ σ3 ∗ ∪ σ4 ∗ σ 1 (τ + i + j/τ) 5. I T T T T T , where 2 . We ﬁnish this section by showing that this is an exhaustive list of all ﬁnite subgroups of S3.

Theorem 17. Any ﬁnite subgroup of S3 SU(2) is either cyclic, or conjugate to ∗ ∗ ∗ ∗ one of the binary subgroups Dn, T , O , or I .

Proof. Let G ⊂ S3 ∼ SU(2) be a finite subgroup with corresponding ∗ 3 subgroup G0 in SU(2)/{±I}. Let G ⊂ S be the inverse image of G0 under the canonical projection SU(2) → SU(2)/{±I}. Clearly, ∗ G ⊂ G . By Theorem 16, G0 is isomorphic to a finite subgroup of ∗ ∗ SO(3).IfG G , then G is the double cover of the group G0.In this case the theorem follows from the classification of finite subgroups in SO(3) (Theorem 11). Thus, we need only study the case ∗ ∗ where G G . In this case G is of index 2 in G , and G and G0 are isomorphic. We first claim that G is of odd order. Assume, to the contrary, that G has even order. By a standard result in group theory (due to Cauchy), G must contain an element of order 2. Since the only element of order 2 in SU(2) is −I, it must be contained in G. But {±I} is the kernel of the canonical projection, and this contradicts G G∗. Thus, G has odd order. G is isomorphic to G0, and as noted above, the latter has an isomorphic copy in SO(3). The odd order subgroups in SO(3) are cyclic, as follows again from the classification of all finite subgroups in SO(3). The theorem follows.

♠ One truly ﬁnal note: In the same way as we derived the real projective plane RP2 as the quotient of S2 by the group {±I},wecan Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

328 23. Back to R3!

consider quotients of S3 by the ﬁnite subgroups above. We obtain 3 3 ∗ the lens spaces L(n; 1) S /Cn, the prism manifolds S /Dn, the tetrahedral manifold S3/T∗, the octahedral manifold S3/O∗, and the icosahedral manifold S3/I∗.

Problems

1. Verify the following inclusions among the binary groups: C ⊂ C ⊂ ∗,C⊂ C ⊂ ∗,D∗ ⊂ ∗; 2 4 T 3 6 T 2 T C ⊂ C ⊂ C ⊂ ∗,D∗ ⊂ D∗ ⊂ ∗,D∗ ⊂ ∗; 2 4 8 O 2 4 O 3 O C ⊂ C ⊂ ∗,C⊂ C ⊂ ∗,C⊂ C ⊂ ∗,D∗,D∗,D∗, ⊂ ∗. 2 4 I 3 6 I 5 10 I 2 3 5 I

2. Verify by explicit calculation that the rotation in H0 represented by the quaternion σ (1/2)(τ + i + j/τ) permutes the vertices of the icosahedron. (What is the rotation angle?) Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Invariants . . . 24 . SECTION .

† ♣ Recall from Section 18 that we set ourselves to the task of finding all finite Mobius¨ groups. Using Cayley’s theorem, we constructed a list of finite Mobius¨ groups most of which arose from the symmetry groups of Platonic solids. To see the difficulty in our quest, we now briy recall from Section 17 the more elementary classification of all finite groups G of spatial rotations. To pin down the structure of G, we considered the set Q of antipodal pairs of poles in S2, the intersections of the axes of the rotations with S2 that constituted G. We showed that Q was G-invariant and that the rotations that corresponded to the poles in a single G-orbit C had the same order. The number of rotations in G with poles in a fixed G-orbit C and of order dC worked out to be (d − )|G| C 1 . 2dC Adding up, we obtained the Diophantine equation |G|− |G| − 1 . 2 2 1 d C∈Q/G C We finally realized that this Diophantine equation was so restrictive that we could simply list all possible scenarios. 329 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

330 24. Invariants

In the more general case when G is a finite subgroup of Mob¨ (Cˆ ), we do not have as much Euclidean structure as for the finite rotation groups, but a simple observation gives a clue how to proceed. Assume that we can find a rational function q : C → C whose invariance group is G, that is, q ◦ g q for g ∈ Mob¨ (Cˆ ) iff g ∈ G. Extended to Cˆ , q can be considered as the projection of an analytic |G|-fold branched covering q : Cˆ → Cˆ (denoted by the same symbol). The branch points are the fixed points of the linear fractional transformations in G, and the branch numbers correspond to the orders of the rotations (minus one) in the special case when G is defined by a rotation group via Cayley’s theorem. By the Riemann–Hurwitz relation (Section 19), the total branching number is 2|G|−2. This will give the same Diophantine restriction for our finite Mobius¨ group G as above! We actually want to prove that the list of finite Mobius¨ groups obtained in Section 18 is exhaustive, that is, any finite Mobius¨ group G is conjugate to one of the groups in that list. To do this, we first construct, for each finite Mobius¨ group in our list, an invariant rational function. This way we obtain a list of rational functions. We then construct an invariant rational function for a general finite Mobius¨ group G. Finally, we compare our list of rational functions with the latter using either uniformization or counting residues. We now see the idea behind this seemingly circuitous argument: A rational function completely characterizes its invariance group, and comparing rational functions is a lot easier than comparing Mobius¨ groups, since to accomplish the former task the entire arsenal of complex analysis is at our disposal!

Remark. There is a quick algebraic way to find all finite Mobius¨ groups G ⊂ Mob¨ (Cˆ ). It is based on the fact that any finite subgroup of SL(2, C) (such as the binary cover G∗ of G) is conjugate to a subgroup of SU(2) (and the finite subgroups of SU(2) are classified in Theorem 17). This is usually proved by averaging the standard scalar product on C2 over the finite subgroup of SL(2, C). Finally, by Cayley’s theorem, a finite subgroup of SU(2) corresponds to a finite group of spatial rotations whose classification was accomplished in Section 17. Despite the existence of this short and elegant proof, we prefer Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

24. Invariants 331 to follow a longer path not only because of its beauty but because we will use some of the ingredients in later developments. The geometry of the spherical Platonic tessellations (including the dihedron) can be conveniently described by the so-called characteristic triangle. Given a spherical Platonic tessellation, a spherical flag (v,e,f) consists of a vertex v, an edge e, and a face f with v ∈ e ⊂ f(⊂ S2). Each spherical flag (v,e,f)contains one characteristic triangle whose vertices are v0 v, v1, the midpoint of the edge e, and v2, the centroid of the face f . (For spherical tessellations we use the spherical analogues of the Euclidean polyhedral concepts such as vertex, edge, face, etc., with obvious meanings.) The spherical angles of a characteristic triangle at the respective vertices are π/ν0, π/ν1, and π/ν2, where ν0,ν1,ν2 are integers greater than or equal to 2. Since v1 is the midpoint of an edge, we always have ν1 2. For the dihedron, we have ν0 2, ν2 n; for the tetrahedron, ν0 ν2 3; for the octahedron, ν0 4, ν2 3; and, finally, for the icosahedron, ν0 5, ν2 3. (Also, ν0 is the number of faces meeting at a vertex, and ν2 is ν0 of the reciprocal; {ν2,ν0} is then the Schli¨ symbol.) ♥ Rections in the sides of a characteristic triangle generate the symmetry group in which the group of direct symmetries of the tessellation is a subgroup of index two. ♣ Each symmetry axis of a spherical Platonic tessellation goes through a vertex, or the midpoint of an edge, or the centroid of a face. It follows that ν0, ν1, and ν2 are the orders of the rotations with axes through the respective points. In particular, the symmetry rotations around the midpoints of edges are always half-turns. The acting symmetry group G of the tessellation has three special orbits on S2. The vertices of the tessellation constitute one special orbit. Since a symmetry rotation around a vertex has order ν0, this orbit consists of |G|/ν0 elements. At the same time, this is the number of vertices of the tessellation. Another special orbit consists of the midpoints of the edges; the orbit consists of |G|/ν1 points, and this is also the number of edges. Finally, the third special orbit consists of the centroids of the faces, or equivalently, the vertices of the reciprocal tessellation. The number of faces is thus |G|/ν2. By Euler’s theorem for convex polyhedra (Section 17), we Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

332 24. Invariants

have |G| |G| |G| − + 2. ν0 ν1 ν2 All other orbits of G are principal, that is, the number of points in the orbit is the order |G| of G. Recall that our intermediate purpose is to construct a rational function q : C → C whose invariance group is a given finite Mobius¨ group G in our list. It is now time to discuss our plan. We first consider the binary group G∗ ⊂ SU(2) associated to G, and study invariance of polynomials F : C2 → C under G∗. We will assume that F is homogeneous (of degree d), that is, d F(tz1,tz2) t F(z1,z2), for all t, z1,z2 ∈ C. We will be able to construct two linearly independent G∗-invariant homogeneous polynomials E, F : C2 → C of the same degree. (The common degree will turn out to be equal to the order |G| of the group G.) Due to homogeneity, the rational quotient E/F will define our analytic function q : Cˆ → Cˆ by

E(z1,z2) z1 q(z) ,z ,z1,z2 ∈ C. F(z1,z2) z2 (The quotient E/F factors through the canonical projection C2 − {0}→CP1, where CP1 is the complex projective line identified 2 ˆ with S C; cf. Section 22.) This is because q(z1,z2) depends only on the homogeneous coordinates of the projective point 1 ∗ [z1 : z2] ∈ CP . Notice, finally, that G -invariance can be defined up to constant multiples since common multiples, of E and F cancel in the ratio E/F.

Remark. q is usually called the fundamental rational function, and the problem of inverting q (to be discussed shortly), the fundamental problem.

We are now ready to get started. For brevity, we call a homogeneous polynomial F : C2 → C a form. Given any subgroup G∗ ⊂ SL(2, C), we say that F is G∗-invariant if there exists a Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

24. Invariants 333

∗ ∗ character χF : G → C −{0}, a homomorphism of G into the multiplicative group C −{0}, such that ∗ F ◦ g χF (g) · F, g ∈ G .

∗ 2 Here g ∈ G acts on the argument (z1,z2) ∈ C by ordinary matrix multiplication. The character χF is uniquely determined by F.We ∗ say that F is an absolute invariant of G if χF 0. In general, F is ∗ ∗ an absolute invariant of the subgroup ker χF ⊂ G .IfG is finite, 1 ∗ then χF maps into the unit circle S ⊂ C −{0} (why?), and the ∗ 1 quotient G / ker χF , being isomorphic to a finite subgroup of S ,is cyclic. According to our plan, for each finite Mobius¨ group G in our list, with corresponding binary Mobius¨ group G∗ ⊂ SU(2), we will exhibit two forms E, F of degree |G| such that E and F are both G∗-invariant and have the same character. The function q defined above is the G-invariant rational function we seek. The cyclic group G Cn does not fit in the general framework, since it is not the symmetry group of a spherical Platonic tessellation. Although it is easy to obtain the general Cn-invariant rational function q : C → C of degree n by inspection, it is instructive to go through the planned procedure in this simple case. We thus seek ∗ ∗ the most general Cn-invariant form F of degree n, where Cn C2n is cyclic. A typical diagonal matrix in SU(2) has diagonal entries a, a−1 1/a, where a ∈ C −{0}. We identify this matrix with the −1 first diagonal element a. It acts on (z1,z2) as (az1,a z2), z1,z2 ∈ C. The character χF is uniquely determined by its value on the generator ω eπi/n, a primitive 2nth root of unity. Since ω2n 1, m χF (ω) must be a 2nth root of unity, that is, χF (ω) ω for some ∗ m 0,...,2n − 1. The condition of Cn-invariance for F reduces to

−1 m F(ωz1,ω z2) ω F(z1,z2), z1,z2 ∈ C. Being homogeneous of degree n, the typical monomial participat- F zj zn−j j ,...,n ing in is 1 2 , where 0 . Substituting this into the equation of invariance, we obtain ω2j ωm+n. Since ω isa2nth root of unity, we have 2n | m + n − 2j. This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

334 24. Invariants

In particular, n|m−2j, or equivalently, m−2j nl for some integer l. Hence, 2n|n(l + 1), and l is odd. Inspecting the ranges of j and m, we see that |m − 2j| < 2n, so that l ±1. We have 2j m ± n, so that m and n have the same parity, and j (m ± n)/2. For l 1, we have m ≥ n, and for l −1, m ≤ n. The corresponding monomials are z(m±n)/2zn−(m±n)/2. 1 2

∗ Two linearly independent Cn-invariant forms of degree n and the same character (the same m) exist iff m n, and in this case, the C∗ zn zn general n-invariant form is a linear combination of 1 and 2 . Since z z1/z2, the general rational function q with invariance group Cn is a quotient of two linearly independent forms. We obtain that the most general Cn-invariant rational function is a linear fractional zn zn zn/zn transformation applied to . Notice, in particular, that 1 2 vanishes at the fixed points 0 and ∞ of the rotations that make up Cn. Analytically, the fixed points are the branch points of q considered as a self-map of Cˆ . (Compare this with the discussion before the proof of the FTA in Section 8!) This last remark gives a clue how to obtain invariant forms in the case of spherical Platonic tessellations. We first discuss the dihedron with the special position given in Section 18. Recall that the vertices of a dihedron are the n points distributed uniformly 2 ˆ along the equator of S with e1 being a vertex. On C, these points n constitute the roots of the equation z 1. Since z z1/z2,we see that the most general degree-n form that vanishes on these zn − zn vertices is a constant multiple of 1 2 . What do we mean by a form F vanishing at a point in Cˆ ? After all, F has two complex arguments, z1 and z2!IfF(z1,z2) 0, then by homogeneity, we also have F(tz1,tz2) 0 for all t ∈ C, so that vanishing of F(z1,z2) is a property of the projective point [z1 : z2] rather than the property of a specific representative (z1,z2). Using more modern terminology, a form F factors through the canonical projection C2 −{0}→CP1 and gives a well-defined function on CP1 Cˆ .Weset zn − zn α(z ,z ) 1 2 . 1 2 2 This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

24. Invariants 335

In a similar vein, the most general degree-n form that vanishes on the midpoints of the edges of the dihedron is a constant multiple zn + zn of 1 2 . We deﬁne zn + zn β(z ,z ) 1 2 . 1 2 2 Finally, the midpoints of the 2 hemispherical faces correspond to 0 and ∞, and we set

γ(z1,z2) z1z2. We say that the forms α, β, and γ belong to the dihedron. As a simple ∗ computation shows, all three forms are Dn-invariant, with χα χβ ∗ ∗ and χγ ±1, with +1 corresponding to the cyclic kernel Cn ⊂ Dn. The forms α, β, and γ are algebraically dependent. In fact, we have α2 − β2 + γn 0.

Notice that the degrees of α, β, and γ are |Dn|/ν0, |Dn|/ν1, and |Dn|/ν2, and consequently, the exponents in the equation above are ν0, ν1, and ν2. It is worthwhile to generalize some of the properties of the forms of the dihedron derived above. In fact, in each of the remaining cases of Platonic tessellations, we will have three forms F0, F1, and F2 that will be said to belong to the tessellation in the sense that F0 vanishes on the projected vertices, F1 vanishes on the projected midpoints of the edges, and F2 vanishes on the projected centroids of the faces.

Remark. In trying to make an up-to-date treatment of the subject, we adopted a number of changes, retaining as much classical terminology as possible. For example, our nonstandard notation for the forms F0, F1, and F2 rects the fact that our Fj, j 0, 1, 2, vanishes on the centroids of the j-dimensional “cells” of the Platonic solid.

Since the zero sets of F0, F1, and F2 are the 3 special orbits of the action of G on Cˆ , these forms are all invariants of the correspond- ∗ ing binary Mobius¨ group G . They have degrees |G|/ν0, |G|/ν1, and Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

336 24. Invariants

ν ν ν |G|/ν F 0 F 1 F 2 2. We now claim that the forms 0 , 1 , and 2 (all of degree |G|) are linearly dependent. To do this, recall that away from the three special orbits of G on S2, all orbits are principal. Given a principal orbit, we can ﬁnd a complex ratio µ0 : µ1 such that ν ν µ F 0 + µ F 1 0 0 1 1 vanishes at one point of this orbit. By invariance, this linear combination vanishes at each point of the orbit. In a ν ν µ µ µ F 1 + µ F 2 similar vein, for another ratio 1 : 2, 1 1 2 2 vanishes on the same orbit. These two linear combinations are polynomials of degree |G|, and both vanish on a principal orbit containing |G| points. It follows that they must be constant multiples of each other. We obtain that

ν ν ν λ F 0 + λ F 1 + λ F 2 0 0 1 1 2 2 0

for some λ0 : λ1 : λ2. Notice that for the dihedron, this reduces to the algebraic relation we just derived for the dihedral forms α, β, and γ. We ﬁnally claim that every G∗-invariant form F can be written as a polynomial in F0, F1, and F2. We proceed by induction with respect to the degree of F and exhibit a polynomial factor of F in F0, F1, and F2. Consider the zero set of F. By invariance of F, this set is G-invariant, the union of some G-orbits. If there is a special orbit among these, then F0, F1,orF2 divides F. Otherwise, there must be a principal orbit on which F vanishes. As above, for a complex ν ν µ µ µ F 0 + µ F 1 ratio 0 : 1, the linear combination 0 0 1 1 vanishes on this principal orbit, and therefore it must be a factor of F. The claim follows. We now consider the tetrahedron. Applying the stereographic projection to the vertices of our tetrahedron, we obtain the points + i − i ± √1 , ± √1 . 3 − 1 3 + 1 A form R of degree 4 that vanishes at these points can be obtained by multiplying out the linear factors: + i 2 − i 2 2 1 2 2 1 2 R(z1,z2) z − √ z z − √ z 1 3 − 1 2 1 3 + 1 2 √ z4 − iz2z2 + z4. 1 2 3 1 2 2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

24. Invariants 337

The vertices of the reciprocal tetrahedron are the negatives of the 2 original. By Problem 1 (b) in Section 7, hN (−p) −1/hN (p), p ∈ S , so that these vertices projected to Cˆ are + i − i ∓ √1 , ∓ √1 . 3 + 1 3 − 1 The corresponding form S of degree 4 is √ S(z ,z ) z4 + iz2z2 + z4. 1 2 1 2 3 1 2 2 The vertices of the tetrahedron and its reciprocal are the two alternate sets of vertices of the circumscribed cube. As a byproduct, we see that the product RS vanishes on the vertices of this cube. Expanding, we obtain RS(z ,z ) z8 + z4z4 + z8. 1 2 1 14 1 2 2 The midpoints of the edges of any one of the tetrahedra are the vertices of the octahedron. Projected to Cˆ , these are

0, ∞, ±1, ±i.

Since z1 vanishes at 0, and z2 vanishes at ∞,aformT of degree 8 that vanishes on the vertices above takes the form T(z ,z ) z z (z2 − z2)(z2 + z2) z z (z4 − z4). 1 2 1 2 1 2 1 2 1 2 1 2 We say that the forms R, T, and S belong to the tetrahedron. Recall that this means that R 0 on the vertices, S 0 on the midpoints of the edges, and T 0 on the centroids of the faces of the tetrahedron. We also see that the degrees of R, T, and S are |T|/ν0, |T|/ν1, and |T|/ν2. The forms R and S are invariants of the binary T∗ χ χ D∗ T tetrahedral group ; R S with kernel 2. The form is an absolute invariant. By the general discussion above, R3, S3, and T2 are linearly dependent. Comparing some of the coefﬁcients, we have √ R3 − 12 3iT2 − S3 0.

It is again convenient to make another observation about invariant forms. Given a G∗-invariant form F, the Hessian Hess (F) This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

338 24. Invariants

deﬁned by

∂2F ∂2F

∂z2 ∂z ∂z (F)(z ,z ) 1 1 2 Hess 1 2 ∂2F ∂2F

∂z ∂z ∂z2 2 1 1 ∗ ∗ is also G -invariant. Furthermore, given two G -invariant forms F0 and F , the Jacobian Jac (F ,F ) deﬁned by 1 0 1 ∂F ∂F 0 0 ∂z ∂z (F ,F )(z ,z ) 1 2 Jac 0 1 1 2 ∂F ∂F 1 1

∂z1 ∂z2 ∗ ∗ is also G -invariant. Thus, once we have a G -invariant form F0,we automatically have two additional G∗-invariant forms: the Hessian F1 Hess (F0) and then the Jacobian F2 Jac (F0,F1). We ﬁrst try this for the tetrahedral form R of degree 4. Since the Hessian Hess (R) is of degree 4, it can only be a linear combination of √R and S. By an easy computation, we see that Hess (R) −48 3iS. The Jacobian of R and S is of degree 6, so that without any computation we see that it must be a constant multiple of T.√ Working out the leading coefﬁcient, we see that Jac (R, S) 32 3iT.

Remark. For the dihedral forms α, β, and γ, we have Hess (α) −Hess (β) −(n2(n − )2/ )γn−2 (γ) (α, β) n2 γn−1 1 4 , Hess 1, and Jac 2 , Jac (α, γ) nβ, Jac (β, γ) nα.

Recall that the octahedral Mobius¨ group contains the tetrahedral Mobius¨ group, so that the same is true for the associated binary groups. Thus the O∗-invariant forms are polynomials of the tetrahedral forms R, S, and T. To obtain the forms that belong to the octahedron, we ﬁrst note that by construction, the octahedral form that vanishes on the vertices of the octahedron is T. It is equally clear that the form RS of degree 8 vanishes on the midpoints of the faces of the octahedron, since these points are nothing but the This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

24. Invariants 339 vertices of the cubic reciprocal. Note also that Hess (T) −25RS. We finally have to find a form of degree 12 that vanishes on the midpoints of the edges of the octahedron. Based on analogy with the tetrahedral forms, this form of degree 12 must be the Jacobian Jac (T, RS) R Jac (T, S) + S Jac (T, R). A simple computation shows that Jac (T, R) −4S2 and Jac (T, S) −4R2. Normalizing, we obtain that the middle octahedral form is R3 + S3 . 2 An explicit expression of this form is obtained by factoring R3 + S3 1 (R + S)(R2 − RS + S2) 2 2 (z4 + z4)(z8 − z4z4 + z8) 1 2 1 34 1 2 2 z12 − z8z4 − z4z8 + z12. 1 33 1 2 33 1 2 2 Summarizing, the three forms that belong to the octahedron are T, (R3 + S3)/2, and RS. They are all absolute invariants of the ∗ binary tetrahedral group T . As octahedral invariants we have χT χ(R3+S3)/2 ±1, while RS is an absolute octahedral invariant. The linear relation among T4, (R3 + S3)2/4, and (RS)3 can be deduced from the previous relation between R, S, and T. Squaring both sides of the equation R3 + S3 √ 6 3iT2, 2 we have 1 1 (R3 − S3)2 (R3 + S3)2 − (RS)3 108T4. 4 4 We arrive at R3 + S3 2 108T4 + − (RS)3 0. 2 We now turn to the final case of the icosahedron. Recall that in Section 18 we determined the vertices of the iscosahedron (in Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

340 24. Invariants

special position) projected to Cˆ by the stereographic projection. An icosahedral form I of degree 12 that vanishes on these projected vertices is

4 4 j 4 j 2 3 I(z1,z2) z1z2 (z1 − ω (ω + ω )z2) (z1 − ω (ω + ω )z2) j 0 j 0 z z (z5 − (ω + ω4)5z5)(z5 − (ω2 + ω3)5z5) 1 2 1 2 1 2 z z (z5 − τ−5z5)(z5 + τ5z5) 1 2 1 2 1 2 z z (z10 + (τ5 − τ−5)z5z5 − z10). 1 2 1 1 2 2

Here we used the identity

4 t5 − 1 (t − ωj) j 0

with the substitutions t z/(ω + ω4) and t z/(ω2 + ω3) (cf. Problem 5 (b) of Section 17). The coefﬁcient τ5 − 1/τ5 is the Lucas number L5 11 (cf. Problem 14 (d) of Section 17). (Another way to see this is to factor ﬁrst as 1 1 1 1 τ5 − τ − τ4 + τ2 + 1 + + , τ5 τ τ2 τ4

and then square the deﬁning relation τ − 1/τ 1 to obtain τ2 + 1/τ2 3, and square again to get τ4 + 1/τ4 7. Adding up, we obtain τ5 − 1/τ5 11.) Summarizing, we arrive at the ﬁrst icosahedral form

I(z ,z ) z z (z10 + z5z5 − z10). 1 1 1 2 1 11 1 2 2

Here I is an absolute invariant of I∗. In fact, all invariants of I∗ are absolute, since I∗ has no proper normal subgroup (cf. Problem 6 of Section 17), and thereby no nontrivial character. (The kernel of a nontrivial character would be a proper normal subgroup.) The Hessian Hess (I) is an absolute invariant of degree 20. Thus, it must vanish on the centroids of the faces of the icosahedron, or equivalently, on the vertices of the reciprocal dodecahedron. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

24. Invariants 341

Normalizing, we compute 1 H(z ,z ) Hess (I)(z ,z ) 1 2 121 1 2 −(z20 + z20) + (z15z5 − z5z15) − z10z10. 1 2 228 1 2 1 2 494 1 2 The Jacobian Jac ( I, H) is an absolute invariant of degree 30, so it must vanish on the midpoints of the edges of the icosahedron. An easy computation shows that 1 J (z ,z ) Jac ( I, H)(z ,z ) 1 2 20 1 2 (z30 + z30) + (z25z5 − z5z25) − (z20z10 + z10z20). 1 2 522 1 2 1 2 10005 1 2 1 2 The icosahedral forms I, J , and H are algebraically dependent. A comparison of coefﬁcients shows that

1728 I5 − J 2 − H3 0.

♠ Let C[z1,z2] denote the ring of polynomials in the variables z1 and z2. Since all icosahedral invariants are absolute, we obtain ∗ that the ring C[z, w]I of icosahedral invariants is isomorphic to the polynomial ring

∗ C[z, w]I C[ I, J , H]/(1728 I5 − J 2 − H3), where we factor by the principal ideal generated by 1728 I5 − J 2 − H3. ♣ We now return to the general situation. Forming the six possi- ν ν ν F 0 F 1 F 2 G ble quotients of 0 , 1 , and 2 , we obtain six -invariant rational functions that solve our problem. Since we have linear relations among these forms, the six rational functions can be written in terms of each other in obvious ways. A more elegant way to express these is to use homogeneous coordinates and write

ν ν ν q q − −λ F 2 λ F 1 λ F 0 , : 1:1 2 2 : 1 1 : 0 0 ν ν q −λ F 2 /λ F 0 where 2 2 0 0 . We make an exception for the dihedron and deﬁne q such that q : q − 1:1 α2 : β2 : −γn. We summarize our results in the following tables: Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

342 24. Invariants

Platonic solid G |G| ν0 ν1 ν2 F0 F1 F2

Dihedron Dn 2n 22nα β γ

Tetrahedron T 12323RT S O T R3+S3 RS Octahedron 24423 2 Icosahedron I 60523IJ H

ν ν ν GλF 0 + λ F 1 + λ F 2 q q − 0 0 1 1 2 2 0 : 1:1

2 2 n 2 2 n Dn α − β + γ 0 α : β : −γ √ √ TR3 − 12 3iT2 − S3 0 S3 : −12 3iT2 : R3 O T4 + R3+S3 2 − (RS)3 (RS)3 R3+S3 2 T4 108 2 0 : 2 :108

I 1728I5 − J 2 − H3 0 H3 : −J 2 : 1728I5

We are ﬁnally ready to prove that any ﬁnite Mobius¨ group is conjugate to one of the Mobius¨ groups listed in Section 18. This result is due to Klein.

Theorem 18. ˆ Any ﬁnite M¨obius group G ⊂ Mob¨ (C) is cyclic, or conjugate to Dn, T, O, or I.

Proof. Let G ⊂ Mob¨ (Cˆ ) be a ﬁnite subgroup. Let a, b ∈ C such that g(a) b, for all g ∈ G. It follows that g(b) a, for all g ∈ G. Consider the rational function q˜ : C → C deﬁned by g(z) − a q(z)˜ ,z∈ . g(z) − b C g∈G Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

24. Invariants 343

The condition on a and b guarantees that q˜ is nonconstant. Given w ∈ C,tofindz ∈ C such that q(z)˜ w, we need to solve the equation (g(z) − a) w (g(z) − b). g∈G g∈G Multiplying out the denominators in the linear fractions g(z), g ∈ G, this becomes a polynomial equation of degree |G| with w as a parameter. Extended to Cˆ , q˜ is an analytic map of degree |G|n the other hand, q˜ is G-invariant. It follows that for fixed w ∈ Cˆ , the solution set of q(z)˜ w is a single G-orbit, and q˜ : Cˆ → Cˆ /G Cˆ is the orbit map. The point z ∈ Cˆ is a branch point iff the G-orbit G(z) through z is not principal. In this case, |G(z)| |G|/ν, and the branch number associated to z is ν − 1. Letting U denote the set of branch points of q˜, the total branching number is |G| B (ν − ), ν 1 U/G where the summation is over all branch values. By the Riemann– Hurwitz relation (Section 19), the total branching number is equal to 2|G|−2, since both the domain and the range have zero genera. We thus have − 1 − 2 . 1 ν 2 |G| U/G As noted at the beginning of this section, this is the same Diophan- tine restriction as the one for the classification of finite rotation groups. Adopting the analysis there in our setting, we see that the sum on the left-hand side either consists of two terms with ν0 ν1 |G| or consists of three tems with ν0, ν1, and ν2 as given in the table above. In the first case, let w0 and w1 be the branch values corresponding to ν0 and ν1. By performing a linear fractional transformation on the range, we may assume that w0 0 and w1 ∞. For the rest of the cases, let w0, w1, and w2 be the three branch values. Performing a linear fractional transformation on the range again, we may assume that these are w0 0, w1 1, and w2 ∞, and they correspond to ν2, ν1, and ν0. (This patterns the zeros and poles of q, since Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

344 24. Invariants

ν ν ν q q − −λ F 2 λ F 1 λ F 0 : 1:1 2 : 1 1 : 0 0 . In the case of the dihedron ν0 and ν2 are switched.) Matching the branch points of q˜ with one of the rational functions q in our list, we arrive at a scenario in which the analytic branched coverings q and q˜ have the same branch points and branch numbers. We now apply a general uniformization theorem for branched coverings1 and conclude that the group G is conjugate to the Mobius¨ group that deﬁnes q and that the conjugation is a linear fractional transformation that establishes the conformal equivalence of the branched coverings q and q˜.

Remark. The use of the powerful uniformization therorem at the end of our proof can be dispensed with. We will give a more elementary approach to the ﬁnal step in the proof above in Section 25/B.

Problem

1. Derive the following table for the absolute invariants of the ﬁnite Mobius¨ groups:

G∗ Absolute invariants Relation

D∗ z2n + z2n z z (z2n − z2n)z2z2 z2n + z2n 2z2z2 − z z (z2n − z2n) 2 z2z2 n+1 n 1 2 1 2 1 2 1 2 [ 1 2 ] 1 2 [ 1 2 1 2 ] 4[ 1 2] 2 T∗ T R3+S3 RS T4 + R3+S3 − (RS)3 2 108 2 0 + , 2 O∗ T2 T R3+S3 RS T2 3 + T R3+S3 − T2 RS 3 2 108[ ] 2 [ ] 0

I∗ IJH 1728I5 − J 2 − H3 0

1See H. Farkas and I. Kra, Riemann surfaces, Springer, 1980. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

. . . . . The Icosahedron and . . . . . the Unsolvable . . . 25 . SECTION . Quintic

† ♠ According to Galois theory1 a polynomial equation is solvable by radicals iff the associated Galois group is solvable. For example, since the alternating group A5 is simple (cf. Problem 6 of Section 17), there is no root formula for a quintic with Galois group A5. Since the symmetry group of the icosahedron is (isomorphic to) A5 (Section 17), the question arises naturally whether there is any connection between the icosahedron and the solutions of quintic equations. This is the subject of Klein’s famous Icosahedron Book.2 We devote this (admittedly long) section to sketch Klein’s main result. We will treat the material here somewhat differently than Klein, and rely more on geometry. Unlike the Icosahedron Book,

1From now on, we will use some basic facts from Galois theory. For a quick summary, see Appendix F. 2The Icosahedron Book first appeared in German: Vorlesungenuber ¨ das Ikosaeder und die Auflösung der Gleichungen vom fünften Grade, Teubner, Leipzig, 1884. Several English editions exist today, for example, Lectures on the icosahedron, and the solution of equations of the fifth degree, Kegan Paul, Trench, Trubner¨ and Co., 1913. A new German edition with the original title (containing various comments and explanations) was published by Birkhauser-Basel,¨ Teubner-Leipzig in 1993. A good summary of the Icosahedron Book is contained in Slodowy’s article, Das Ikosaeder und die Gleichungen fünften Grades, in Mathematischen Miniaturen,¨ Band 3, Birkhauser-Basel,¨ 1986. For a recent easy-to-follow text, see J. Shurman, Geometry of the Quintic, John Wiley & Sons, 1997. 345 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

346 25. The Icosahedron and the Unsolvable Quintic

we will take as direct a path to the core results as possible. Due to the complexity of the exposition, we divide our treatment into subsections.

A. Polyhedral Equations

In Section 24, we deﬁned, for each ﬁnite Mobius¨ group G,a G-invariant rational function q : C → C. Geometrically, the extension q : Cˆ → Cˆ is the analytic projection of a branched covering between Riemann spheres, and the branch values are w 0, w 1, and w ∞, with branch numbers ν2, ν1, and ν0 minus one. (In the case of the dihedron ν0 and ν2 are switched.) As in the proof of Theorem 18, for a given w ∈ C, the equation q(z) w for z can be written as a degree-|G| polynomial equation

P(z) − wQ(z) 0,

where q P/Q with P and Q the polynomial numerator and denominator of q (with no common factors). We call this the polyhedral equation associated to G. Clearly, this polynomial equation has |G| solutions (counted with multiplicity, and depending on the parameter w). We now consider this equation for each G. The case of the cyclic group Cn is obvious, since the associated equation is zn w, and the solutions are simply the nth roots of w. Using the second table in Section 24, for the equation of the dihedron we have α(z ,z )2 (zn − zn)2 (zn − 1)2 q (z) − 1 2 − 1 2 − w, Dn γ(z ,z )n znzn zn 1 2 4 1 2 4 where we have indicated the acting group by a subscript. Multi- plying out, we obtain the equation of the dihedron, a quadratic equation in zn. This can easily be solved: z q−1(w) n − w ± w(w − ). Dn 1 2 2 1 q Inverting Dn amounts to extracting a square root followed by the extraction of an nth root. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

A. Polyhedral Equations 347

For the tetrahedron, we have √ S(z ,z )3 z4 + iz2z2 + z4 3 1 2 1 2 3 1 2 2 qT (z) √ R(z ,z )3 z4 − iz2z2 + z4 1 2 1 2 3 1 2 2 √ z4 + iz2 + 3 2√3 1 w. z4 − 2 3iz2 + 1 Taking the cube root of both sides, we arrive at an equation that is 2 quadratic in z and can be easily solved. Inverting qT thus amounts to extracting a cube root followed by the extraction of two square roots. Comparing the expression of qT just obtained with that of q d Dn for 2, we have q (z) − eπi/3 3 D2 qT(z) . q (z) + e2πi/3 D2 As noted in Section 24, the octahedral invariants can be written as polynomials in the tetrahedral invariants. We have q (z) (RS)3 (S/R)3 w O , qO(z) − 1 ((R3 + S3)/2)2 ((S/R)3 + 1)/2)2 w − 1 where we have omitted the arguments z1 and z2 for simplicity. Multiplying out, we obtain a quadratic equation in (S/R)3. Since this is qT(z), we see that inverting qO amounts to extracting a square root followed by the extraction of a cube root, and followed by the extraction of two square roots.

Remark. In view of Galois theory, it is illuminating to match the sequence of −1 roots in the root formulas for qG with the indices of the consecutive normal subgroups in a composition series of G, where G is one of our (solvable) groups G Cn,Dn,T,O. For example, a composition series for O is S4 ⊃ A4 ⊃ D2 ⊃ C2, and the sequence of indices is 2, 3, 2, 2!

The situation for the icosahedron is radically different. We have H3 qI (z) w. 1728 I5 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

348 25. The Icosahedron and the Unsolvable Quintic

As we will see later, it is impossible to express the solutions of the icosahedral equation H3(z, 1) − 1728w I5(z, 1) 0 in terms of a radical formula (depending on w). Using the explicit expressions of the forms involved, the equation of the icosahedron can be written as ((z20 + 1) − 228(z15 − z5) + 494z10)3 + 1728wz5(z10 + 11z5 − 1)5 0.

B. Hypergeometric Functions

We just noted that there is no general root formula for the solutions of the icosahedral equation. The question arises naturally as to what kind of additional “transcendental” procedure is needed to express the solutions in an explicit form. In this subsection we show that any solution of a polyhedral equation can be written as the quotient of two linearly independent solutions of a homogeneous second-order linear differential equation with exactly 3 singular points, all regular. These differential equations are called hypergeometric.3 Equivalently, we will show that for each spherical Platonic tessellation, the inverse q−1 of the rational function q is the quotient of two hypergeometric functions.

Remark. Now some history. Bring (in 1786) and Jerrard (in 1834) independently showed that the general quintic can be reduced to z5 + bz + c (by a suitable Tschirnhaus tranformation; cf. the next subsection). This is usually called the Bring–Jerrard form. By scaling, the Bring–Jerrard form can be further reduced to the special z5 quintic + z − c 0. A root√ of this polynomial is called an ultraradical, and it is denoted by ∗ c. Using the deﬁning equation, an ultraradical can be easily expanded into a convergent series. Bring and Jerrard thus showed that the general quintic can be solved by the use of radicals and ultraradicals. The relation of this special

3From now on, we use the deﬁnitions and results of Appendix E without making further references. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

B. Hypergeometric Functions 349 quintic to the so-called modular equation was used by Hermite, who pointed out that the general quintic can be solved in terms of elliptic modular functions.

As usual, we let G ⊂ Mob¨ (Cˆ ) denote the invariance group of q. Recall that q is an analytic branched covering with branch values w0 0, w1 1, and w2 ∞, and branch numbers ν2, ν1, and −1 ν0 minus one. The inverse q is multiple-valued. By G-invariance, composing q−1 with an element of G allows us to pass from a single- valued branch of q−1 to another. Since the Schwarzian is invariant under any linear fractional transformation, S(q−1) must be single- valued:

S(q−1) s, where s : C → C is a rational function. Following Riemann, who was a strong proponent of the principle that a rational function is best described by its poles, we determine s explicitly by taking its Laurent expansion at its poles. Near a branch value wj, j 0, 1, 2, q−1 can be expanded locally as

−1 −1 1/ν2−j 2/ν2−j q (w) − q (wj) a1(w − wj) + a2(w − wj) +···,

−1 where q (wj) denotes any one of the |G|/ν2−j preimages, and −1 −1 q (w) is near q (wj). As usual in complex analysis, we agree that for wj ∞, w −wj means 1/w. Substituting this expansion into the expression for the Schwarzian S, we obtain that the initial terms of the series for s at w0 0, w1 1, and w2 ∞are ν2 − 1 ν2 − 1 ν2 − 1 2 , 1 , 0 . ν2w2 ν2(w − )2 ν2w2 2 2 2 1 1 2 0

(In the case of the dihedron, ν0 and ν2 are switched.) We obtain ν2 − 1 A ν2 − 1 B s(w) 2 + + 1 + + C, ν2w2 w ν2(w − )2 w − 2 2 2 1 1 1 where A, B, C are complex constants. These constants are determined by the behavior of s at inﬁnity, namely, by the requirement that s, expanded into a Laurent series at ∞, have the initial term Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

350 25. The Icosahedron and the Unsolvable Quintic

(ν2 − )/ ν2w2 A + B C 0 1 2 0 . We ﬁnd that 0, 0, and ν2 − 1 ν2 − 1 ν2 − 1 2 + 1 + B 0 . ν2 ν2 ν2 2 2 2 1 2 0 Putting all these together, we ﬁnally arrive at

1 1 1 2 2 + − − ν − 1 ν − 1 ν2 ν2 ν2 1 s(w) 2 + 1 + 1 2 0 . ν2w2 ν2(w − )2 w(w − ) 2 2 2 1 1 2 1

Remark. Recall that in the proof of Theorem 18 we constructed, for a given ﬁnite Mobius¨ group G, a rational function q˜ that had the same branch points and branch numbers as q. Once again, passing from a single-valued branch of q˜ −1 to another amounts to the composition of q˜−1 with a linear fractional transformation in G. We thus have S (q˜ −1) s˜, where s˜ is a rational function. On the other hand, since the branch points and branch numbers of q and q˜ are the same, the singularities of s and s˜ are also the same. Since a rational function is determined by its singularities, we must have s s˜.We obtain S(q−1) S(q˜−1), so that q and q˜ differ by a linear fractional transformation. This linear fractional transformation conjugates G to the ﬁnite Mobius¨ group corresponding to q. Thus, Theorem 18 follows.

Returning to the main line, we see that q−1 satisﬁes the third- order differential equation

2 2 1 + 1 − 1 − ν − 1 ν − 1 ν2 ν2 ν2 1 S(q−1) 2 + 1 + 1 2 0 . ν2w2 ν2(w − )2 w(w − ) 2 2 2 1 1 2 1 Although of third order, the general solution of this equation is remarkably simple and can be given in terms of solutions of the homogeneous second-order linear differential equation z p(w)z + q(w)z,

where 1 s p − p2 − 2q. 2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

C. The Tschirnhaus Transformation 351

We set p(w) 1/w, and choose q to satisfy the equation for s. Substituting the actual expression of s into p and q, we obtain z z 1 1 1 1 w2 z + + − + w + − + 1 − 0. w (w − )2w2 ν2 ν2 ν2 ν2 ν2 4 1 2 0 2 1 0 This is a special case of the hypergeometric differential equation.In fact, with ν2 − 1 1 1 1 1 α1 −α2 ,β1 ,β2 ,γ1 −γ2 , 2ν2 2ν1 2ν1 2ν0 the hypergeometric differential equation reduces to our equation, since ν1 2. We have accomplished our goal and proved that the function q−1 can be written as the quotient of two hypergeometric functions!

C. The Tschirnhaus Transformation

In this subsection, we will reduce the general irreducible quintic

5 4 3 2 z + a1z + a2z + a3z + a4z + a5 0 to a simpler form

5 4 3 2 z + a˜ 1z + a˜ 2z + a˜ 3z + a˜4z + a˜ 5 0, in which some (but not all) coefﬁcients vanish. This reduction is made possible by the Tschirnhaus transformation. It is given by

4 (l) z˜ λlz , l 1 where 5 (l) l 1 l z z − zj , 5 j 1 and z1,...,z5 are the roots of the original quintic. In the expression of z(l), the sum of powers is a symmetric polynomial in the roots, and thus, by the fundamental theorem on symmetric polynomials, Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

352 25. The Icosahedron and the Unsolvable Quintic

it can be expressed as a polynomial in the coefﬁcients a1,...,a5. For example, since

5 5 z −a , z2 a2 − a , j 1 j 1 2 2 j 1 j 1 we have a 1 z(1) z + 1 ,z(2) z2 − a2 − 2a . 5 5 1 2 Hence, z˜ is a polynomial in z of degree less than or equal to 4 with coefficients in Q [a1,...,a5] depending on λ1,...,λ4. The requirement on the vanishing of the prescribed coefficients in the reduced quintic amounts to polynomial relations of degree less than or equal to 4 in the coefficients λ1,...,λ4. These polynomial relations are solvable by explicit root formulas (cf. Section 6). The way the Tschirnhaus transformation z˜ acts on the original quintic is to transform its roots z1,...,z5 to the roots z˜1,...,z˜5 of the reduced quintic, where

4 (l) z˜j λlzj ,j 1,...,5. l 1

5 z˜ 4 λ 5 z(l) a˜ Since j 1 j l 1 l j 1 j 0, we have 1 0 for any Tschirnhaus transformation. The simplest Tschirnhaus transformation is a z˜ z(1) z + 1 , 5

where we put λ1 1, λ2 λ3 λ4 0. (Note that the analogue of this for cubics and quartics was used in Section 6.) Next, we look for a Tschirnhaus transformation that yields a˜ 2 0 in the form a 1 z˜ λz(1) + z(2) λ z + 1 + z2 − a2 − 2a , 5 5 1 2 λ ∈ λ λ where C is a parameter to be determined. Here we set 1, 5 λ 1, and λ λ 0. Since z˜j 0, by squaring we see 2 3 4 j 1 a˜ 5 z˜2 that the vanishing of 2 amounts to the vanishing of j 1 j . This Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

C. The Tschirnhaus Transformation 353 gives the following quadratic equation for λ:

5 5 2 (1) (2) 2 z˜j (λzj + zj ) j 1 j 1 5 5 5 2 (1) 2 (1) (2) (2) 2 λ zj + 2λ zj zj + zj 0. j 1 j 1 j 1

Again, by the fundamental theorem on symmetric polynomials, the coefﬁcients of this quadratic polynomial in λ depend only on a1,...,a5. The corresponding quadratic equation can be solved for λ in terms of a1,...,a5. The two solutions for λ involve the square root of the expression 5 2 5 5 (1) (2) (1) 2 (2) 2 4 zj zj − 4 zj zj . j 1 j 1 j 1

δ ( 5 (z(1))2)2 This is the discriminant multiplied by j 1 j . The smallest ground field over which the original quintic√ is defined is k (a ,...,a ) δ ∈ k δ/∈ k Q 1 5 . We have , but√ in general, . Thus, if we want λ to be in the ground field, δ needs to be adjoined to k. Summarizing (and adjusting the notation), the problem of solvability of the general quintic is reduced (at the expense of a quadratic extension of the ground field) to solvability of the equation

P(z) z5 + 5az2 + 5bz + c 0.

(Here we have inserted numerical factors for future convenience.) A quintic with vanishing terms of degree 3 and 4 (such as P above) is said to be canonical. For future reference we include here the discriminant 2 δ (zj − zl) 1≤j

354 25. The Icosahedron and the Unsolvable Quintic

This formula is obtained by a somewhat tedious but elementary computation.4 Consider the roots z1,...,z5 of the canonical equation as homogeneous coordinates of a point [z1 : ··· : z5] in the 4-dimensional complex projective space CP4 (cf. Section 21). (Here the trivial case 5 P(z) z needs to be excluded, since it gives z1 ··· z5 0.) The roots z1, ···,z5 can be reconstructed from the projective point [z1 : ··· : z5] by extracting a square and a cube root, so that no information is lost in the homogenization. Since ordering the roots cannot be prescribed universally, to the roots there correspond 120 projective points obtained from one another by permuting the coordinates. To incorporate this ambiguity, we consider the sym- 4 metric group S5 acting on CP by permuting the homogeneous coordinates. In other words, the 120 points form an orbit in CP4 S S under- the action of. 5. Notice that under the action of 5, the z(l) ··· z(l) l ,..., points 1 : : 5 , 1 4, are permuted the same way as [z1 : ··· : z5]. Since our equation is canonical, the roots z1,...,z5, satisfy the relations

5 5 2 zj 0 and zj 0. j 1 j 1

Thus, [z1 : ··· : z5] lies in the (smooth) complex surface ( ) 5 5 4 2 Q0 [z1 : ··· : z5] ∈ CP | zj zj 0 . j 1 j 1

Actually, Q0 can be identiﬁed with the so-called standard complex 3 projective quadric in CP , once we identify the complex projective P3 P3 P4 5 z space C with the linear slice C 0 of C deﬁned by j 1 j 0. We thus set ( ) 5 P3 z ··· z ∈ P4 | z . C 0 [ 1 : : 5] C j 0 j 1

4I must confess the advantage in using a computer algebra system such as Maple or Mathematica that reduces the computation of δ to a fraction of a second. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

D. Quintic Resolvents of the Icosahedral Equation 355

The geometry behind the Tschirnhaus transformation that reduces the general quintic to a canonical form is now clear. We start with 4 the point [z1 : ··· : z5] ∈ CP whose homogeneous coordinates are the roots of a general quintic (with no quartic term). We form z(1) ··· z(1) z(2) ··· z(2) the projective points [ 1 : : 5 ] and [ 1 : : 5 ] that both P3 lie in C 0 . We then consider one of the two intersections of the projective line through these two points with the quadric Q0. The Tschirnhaus transformation associates to [z1 : ··· : z5] this intersection point in Q0. As emphasized above, this process generally requires a quadratic extension of the ground ﬁeld k.

D. Quintic Resolvents of the Icosahedral Equation

Let k be our ground field. For the sake of concreteness, we assume that k ⊂ C. Since k has characteristic zero, we also have Q ⊂ k. Since the icosahedral equation contains an arbitrary parameter w, it is natural to consider this equation to be defined over k(w), the field of rational functions in the variable w and with coefficients in k. Let z denote a solution of the icosahedral equation. Since q(z) w, the field k(z) contains k(w). Assume from now on that ω e2πi/5, a primitive fifth root of unity, is contained in k, so that Q (ω) ⊂ k ⊂ C. This ensures that the linear fractional transformations in the icosahedral Mobius¨ group I A5 are defined over k. They act, by substitutions, as automorphisms of the field k(z). (More precisely, g ∈ I acts on r∗ ∈ k(z) by g : r∗ → r∗ ◦ g−1.) Since q is A5-invariant, these automorphisms fix the subfield k(w). ˆ ˆ A5 In fact, since q : C → C is the orbit map of A5, the fixed field k(z) is k(w). We obtain that k(z)/k(w) is a Galois extension with Galois group A5. Moreover, k(z) is the splitting field of the icosahedral equation over k(w); the solutions are the linear fractions that represent the transformations in the icosahedral Mobius¨ group. Since A5 is transitive on the solutions, the icosahedral equation is irreducible over k(w). As before, a rational function r∗ ∈ k(z) has the Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

356 25. The Icosahedron and the Unsolvable Quintic

resolvent polynomial

n∗ ∗ ∗ P (X) (X − rj ), j 1 k(w) A (r∗) {r∗,...,r∗ } with coefﬁcients in , where 5 1 n∗ is the orbit r∗ I k(z) k(r∗,...,r∗ ) through . Since is simple, 1 n∗ . The rest of this subsection is going to be very technical. In order to see through the details we now sketch our plan. Recall that our main objective is to establish a connection between the (irreducible) quintic in canonical form

P(z) z5 + 5az2 + 5bz + c

and the solutions of the icosahedral equation

H3(z, 1) − 1728w I5(z, 1) 0.

Since the latter is a polynomial equation of degree 60, based on the analogy with the cubic resolvent of quartics (cf. Section 6), we must find a suitable quintic resolvent of H3 − 1728w I5 directly comparable to our canonical quintic. Our task is actually harder than in the quartic case, since P depends on three complex coefficients.5 Thus, we need to find a quintic resolvent in canonical form that contains, in addition to w, two extra parameters, u and v, say. We write this resolvent as

P∗(z) z5 + 5a(u, v, w)z2 + 5b(u, v, w)z + c(u, v, w).

We expect the coefﬁcients to be explicitly computable rational functions in u, v, w. Since P∗ is an icosahedral resolvent, we also need to be able to express the roots of P∗ in terms of the solutions of the icosahedral equation, or, what is the same, in terms of hypergeometric functions. Once P∗ is worked out explicitly, the matching with the irreducible quintic P with coefﬁcients a, b, c amounts to solving the

5Shurman’s approach in his Geometry of the Quintic cited above is different (see also Problem 4). Both approaches are contained in the Icosahedron Book. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

D. Quintic Resolvents of the Icosahedral Equation 357 system a(u, v, w) a, b(u, v, w) b, c(u, v, w) c.

This we will be able to carry out with the additional constraint √ δ(u, v, w) δ, where δ(u, v, w) is the discriminant of P∗. This matching process will establish that the roots of P and P∗ coincide as sets. To obtain a root-by-root match we will also need to look at how the Galois group A5 acts on each set of roots. The five roots of a quintic resolvent constitute an orbit of A5 in the extension k(z)/k(w). We thus need to find five rational functions in z that are permuted among themselves by the icosahedral substitutions. During the construction of the rational function q we learned that it is much easier to construct forms first. Our task thus reduces to finding an A5-orbit of five forms. A further advantage in using forms is that we can derive them from geometric situations. An obvious example for five geometric objects that are permuted among themselves by the symmetries of the icosahedron are Kepler’s five cubes inscribed in the reciprocal dodecahedron (cf. Section 17). We could immediately construct five degree-8 forms that vanish at the vertices of these cubes. This would give us a quintic resolvent. For technical purposes, however, it is more convenient to construct the five degree-6 forms that vanish on the vertices of the octahedral reciprocals. Thus, we stay with (Pacioli’s model of) the icosahedron and notice that the 6 midpoints of the 6 shorter edges of the three mutually perpendicular golden rectangles give the 6 vertices of an octahedron. There are 5 configurations each consisting of three mutually perpendicular golden rectangles. (The 15 opposite pairs of edges of the icosahedron give 15 golden rectangles inscribed in the icosahedron. By further grouping, these 15 golden rectangles form the 5 configurations.) Toeach configuration we can attach an octahedron as above. Putting these together, we arrive at the compound of 5 octahedra Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

358 25. The Icosahedron and the Unsolvable Quintic

inscribed in the icosahedron. The octahedral forms corresponding to the five octahedra will be the roots of our quintic resolvent. To choose the first octahedron, we recall that the commuting half-turns U and V introduced in Section 18 have orthogonal axes, so that the composition W UV is also a half-turn and its axis is orthogonal to those of U and V. The linear fractional transformation that corresponds to U is z →−1/z, while those that correspond to V and W have been explicitly worked out in Section 18. We choose the first octahedron (projected to Cˆ ) to have vertices as the fixed points of these half-turns. The fixed points of U are the solutions z2 − z2 + z2 of 1, and the form that vanishes at these points is 1 2, z1,z2 ∈ C, z z1/z2. Using the explicit form for V, for the fixed points of V we need to consider

(ω2 − ω3)z + (ω − ω4)z z 1 2 1 . 4 2 3 (ω − ω )z1 − (ω − ω )z2 z2 Multiplying out, we have

z2 − (ω + ω4)z z − z2 . 1 2 1 2 2 0 By Problem 5 (b) of Section 17, ω + ω4 can be replaced by 1/τ, where τ is the golden section. We obtain that the ﬁxed points of V are given by the zeros of the quadratic form 2 z2 − z z − z2. 1 τ 1 2 2 In a similar vein, the ﬁxed points of W are given by the zeros of

z2 + τz z − z2. 1 2 1 2 2

We deﬁne the ﬁrst octahedral form T1 as the product of these three forms:

T (z ,z ) (z2 + z2)(z4 + z3z − z2z2 − z z3 + z4). 1 1 2 1 2 1 2 1 2 6 1 2 2 1 2 2

The remaining four octahedral forms Tj+1, j 1,...,4, are obtained by applying to T1 the homogeneous substitutions

3j 2j z1 →±ω z1,z2 →±ω z2, Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

D. Quintic Resolvents of the Icosahedral Equation 359 which correspond to the rotations Sj, j 1,...,4, introduced in Section 18. Multiplying out, we arrive at the ﬁve octahedral forms T (z ,z ) ω3jz6 + ω2jz5z − ωjz4z2 j+1 1 2 1 2 1 2 5 1 2 − ω4jz2z4 − ω3jz z5 + ω2jz6,j ,..., . 5 1 2 2 1 2 2 0 4

How does A5 act on {T1,...,T5}? It is enough to see how the icosahedral generators S and W act on these octahedral forms. By definition, S permutes Tj, j 1,...,5, cyclically. Geometrically, the midpoints of the 5 edges that meet at the North Pole N are vertices of the 5 distinct octahedra. Since the rotation axis of S passes through N, S must permute these vertices cyclically. It is equally clear that U, V, W( UV) all fix T1, since T1 is defined by the requirement that it should vanish at the fixed points of U, V, and W. The simplest of these three commuting Mobius¨ tranformations is U : z →−1/z. The corresponding homogeneous substitution is z1 → z2, z2 →−z1. Substituting this into the explicit expression of the octahedral forms above and looking at the leading coefficients, we see immediately that U acts as follows:

U : T1 → T1,T2 ↔ T5,T3 ↔ T4. Recall from Section 18 that the axis of the half-turn V is perpendicular to the axis of U, and it passes through the midpoint of the base of the upper pentagonal pyramid U of the icosahedron. The vertex of U is N. As noted above, the midpoints of the 5 edges of U that meet at N belong to the 5 inscribed octahedra. Since S permutes these edges cyclically, we denote them by e1,e2,e3,e4,e5, where S(ej) ej(mod 5)+1, j 1,...,5. We adjust the index such that e1 is an edge of a golden rectangle that defines T1. Then, for all j 1,...,5, ej is an edge of a golden rectangle that defines Tj. Due to this arrangement the fixed point of V is the midpoint of the base of the triangular face with sides e3 and e4. Now a careful look at how V acts on e2,e3,e4,e5 reveals that V(e3) and e5 are edges of two golden rectangles in the same configuration. Similarly, V(e4) and e2 are edges of two golden rectangles in another configration. Since V is a half-turn, we obtain that it acts on the five octahedra as follows:

V : T1 → T1,T2 ↔ T4,T3 ↔ T5. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

360 25. The Icosahedron and the Unsolvable Quintic

Finally, the way the permutation W acts follows, since W UV. For future reference, we summarize the actions of S and W as follows:

S : Tj → Tj(mod 5)+1,j 1,...,5,

W : T1 → T1,T2 ↔ T3,T4 ↔ T5.

Despite their simplicity, these octahedral forms will not be suit- 5 T2 able for our purposes, since j 1 j 0, so that the quintic icosahedral resolvent constructed from these forms will not be canonical. (Nevertheless, this resolvent is of great interest, since 5 T 5 T3 j 1 j 0 and j 1 j 0; see Problem 4.) This temporary setback is easy to ﬁx. Consider the Hessians Vj Hess (Tj), j 1,...,5, that are forms of degree 8, and also constitute an A5-orbit. They clearly satisfy the relations 5 5 2 Vj 0, Vj 0, j 1 j 1 since there are no icosahedral forms in degrees 8 and 16. (Any icosahedral form is a polynomial in I, J , and H; cf. Section 24.) Thus, the quintic icosahedral resolvent constructed from these forms will be canonical! For future reference we include here the explicit form V (z ,z ) −ω4jz8 + ω3jz7z − ω2jz6z2 − ωjz5z3 j+1 1 2 1 1 2 7 1 2 7 1 2 + ω4jz3z5 − ω3jz2z6 − ω2jz z7 − ωjz8 7 1 2 7 1 2 1 2 2 (ω4jz − ω3jz )(−z7 + z2z5) 1 2 1 7 1 2 + (ω2jz − ωjz )(− z5z2 − z7), j ,..., . 1 2 7 1 2 2 0 4 Recall that we need a two-parameter family of these forms. Notice that we have 5 5 5 2 2 TjVj 0, TjVj 0, (TjVj) 0, j 1 j 1 j 1 since left-hand sides are icosahedral invariants of degrees 14, 22, and 28, respectively, and (once again) there are no icosahedral invariants in these degrees. For future reference, we include here Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

D. Quintic Resolvents of the Icosahedral Equation 361 the explicit forms

Tj+1(z1,z2)Vj+1(z1,z2) (ω4jz − ω3jz )(− z10z3 + z5z8 + z13) 1 2 26 1 2 39 1 2 2 + (ω2jz − ωjz )(−z13 + z8z5 + z3z10), j ,..., . 1 2 1 39 1 2 26 1 2 0 4 For greater generality, we seeek a resolvent for the linear combinations

ϒj EVj + FTjVj,j 1,...,5, where by homogeneity, the coefﬁcients E and F are forms of degree 30 and 24. In fact, we will put E 12u J ,F 144v I2, where u and v are complex parameters, but for the time being, we keep E and F arbitrary. It remains to work out the resolvent explicitly. Staying in the realm of forms for a while, we write the resolvent polynomial as 5 5 4 3 2 (X − ϒj) X + b1X + b2X + b3X + b4X + b5, j 1 where we used X as variable. By what we said above, the relations above imply that b1 b2 0. Expanding, and using the fact that the coefﬁcients must be (absolute) invariants of the icosahedral group, after somewhat tedious computations, we arrive at the following: X 5 + 5X 2(8E3I2 + E2FJ + 72EF 2I3 + F 3IJ) + 5X (−E4IH + 18E2F 2I2H + EF 3HJ + 27F 4I3H) + (E5H2 − 10E3F 2IH2 + 45EF 4I2H2 + F 5JH2). We introduce the new variable I X X JH along with I2 I r T ,s V ,j ,..., . j 12 J j j 12 H j 1 5 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

362 25. The Icosahedron and the Unsolvable Quintic

Comparing the degrees of the forms involved we see that rj and sj are rational functions in z z1/z2. Rewriting our variables in terms of these, we obtain JH ϒ t ,j ,..., , j I j 1 5 where

tj usj + vrjsj,j 1,...,5. In terms of these new variables, our resolvent polynomial becomes 5 ∗ P (X) (X − tj) j 1 5X2 6uv2 + v3 X5 + 8u3 + 12u2v + w 1 − w 15X 6u2v2 + 4uv3 3v4 + −4u4 + + w 1 − w 4(1 − w)2 3 40u3v2 15uv4 + 4v5 + 48u5 − + . w 1 − w (1 − w)2 This is called the canonical resolvent polynomial of the icosahedral equation. By homogeneity, the roots tj tj(u,v,z) usj(z) + ∗ vrj(z)sj(z), j 1,...,5, of the resolvent P are rational functions in z z1/z2 and linear functions in u and v. As in the case of the octahedral forms, the icosahedral generators S and W induce the following permutations on the roots:

S : tj → tj( mod 5)+1,j 1,...,5,

W : t1 → t1,t2 ↔ t3,t4 ↔ t5. As in the case of the canonical quintic, the projective points corresponding to the roots of the canonical resolvent lie in the complex projective quadric ( ) 5 5 4 2 Q0 [t1 : ... : t5] ∈ CP | tj tj 0 . j 1 j 1 Finally, since the solution set contains two parameters, it is reasonable to expect that the projective points corresponding to the roots Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

E. Solvability of the Quintic a` la Klein 363 of the canonical resolvent ﬁll Q0. This will follow as a byproduct of stronger results in Subsection G. Summarizing, we have found that the complex projective Q P3(⊂ P4) quadric 0 in C 0 C serves as a simple “parameter space” for the solutions of the canonical resolvent of the icosahedral equation. Since the solutions of the canonical resolvent depend rationally on the solutions of the icosahedral equation, our remaining task is to ﬁnd a matching parametrization of the solutions of the quintic in canonical form. This we will accomplish in the remaining three subsections.

E. Solvability of the Quintic a` la Klein

In view of the fact that zn − w 0 is the “polyhedral equation” for the cyclic group Cn, it is natural to look for a noncommutative analogue of the theorem in Appendix F,in which the field extension K/k is generated by any solution of the equation zn − w 0, and the Galois group of the field extension is Cn. In order that the linear fractional transformations that make up the icosahedral Mobius¨ group I become k-automorphisms of the splitting field of the icosahedral equation, we need to assume that ω ∈ k, where ω e2πi/5.

Theorem 19. Let k be a ﬁeld satisfying Q (ω) ⊂ k ⊂ C, and let K ⊂ C be a Galois extension of k with Galois group A5. Then, replacing k by a suitable quadratic extension, there exists w∗ ∈ k such that K is generated by any solution z∗ of the icosahedral equation with parameter w∗ q(z∗). ∗ Moreover, each solution z gives rise to an isomorphism ϕ:A5 → I of the Galois group A5 to the icosahedral M¨obius group I such that if σ ∈ A5 is a k-automorphism of K that is mapped under this isomorphism to

a(σ) b(σ) ϕ(σ) ± , c(σ) d(σ) Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

364 25. The Icosahedron and the Unsolvable Quintic

then a(σ)z∗ + b(σ) σ−1(z∗) ϕ(σ)(z∗) . c(σ)z∗ + d(σ)

Remark. Theorem 19, the so called “Normalformsatz,” is the cornerstone of Klein’s theory of the icosahedron. In 1861, Kronecker showed that the “suitable quadratic extension” k/k (“akzessorische Irra- tionalitat”¨ as Klein called it) in the Normalformsatz cannot, in general, be dispensed with. As shown in the text, this extension appears in reducing the general quintic to a canonical form (called “Hauptgleichung” by Klein) by a Tschirnhaus transformation.

Remark. Quadratic extensions do not change the setting in Theorem 19. In fact, if k is a quadratic extension of the ground ﬁeld k, then k is not contained in K, since the Galois group A5 of the extension K/k cannot contain any subgroup of index 2. We thus have G(K ·k/k) G(K/k) A5.

Every irreducible quintic 5 4 3 2 P(z) z + a1z + a2z + a3z + a4z + a5,a1,...,a5 ∈ C,

over k, Q (ω) ⊂ k ⊂ C, with Galois group A5 has a splitting field K as in Theorem 19. Conversely, given K/k as in Theorem 19, there exists an irreducible quintic over k whose splitting field is K, and whose Galois group is A5. Indeed, consider a subgroup of A5 isomorphic to A4. By abuse of notation, we denote this subgroup A4 by A4. The field k is properly contained in the fixed field K , A z ∈ K 4 − k A z so that there exists 1 . The 5-orbit of 1 consists/ of 5 z ,...,z A A P(z) 5 (z − elements 1 5, since 4 is maximal in 5. Let j 1 zj) be the quintic resolvent associated to z1. Then P is irreducible over k, K is the splitting field of P over k, and the Galois group A5 can be identified with the group of even permutations of the roots z1,...,z5. To prove Theorem 19, we will view K as the splitting field of an A irreducible quintic√ (with Galois group 5). The “suitable quadratic extension” k( δ) of the ground field k is due to the reduction of Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

F. Geometry of the Canonical Equation: General Considerations 365 the quintic to canonical form by a Tschirnhaus transformation as discussed in Subsection C.

F. Geometry of the Canonical Equation: General Considerations

We saw in Subsections D and E that the complex projective quadric Q0 parametrizes the points that correspond to the roots of the canonical resolvent of the icosahedral equation, and to the roots of the irreducible quintic in canonical form (obtained from the general quintic by a Tschirnhaus transformation). In the two subsections that follow, by “matching” these two parametrizations we obtain a constructive proof of Theorem 19 as well as the explicit formulas that solve the canonical quintics. In this subsection we will follow the main argument with few technical details. In the last subsection we will work out all formulas in detail. Q ⊂ P3 The geometry of the projective quadric 0 C 0 as a “doubly ruled surface” is well known. In fact, the so-called Lagrange sub- P3 P3 stitution, a linear equivalence between C 0 and C , transforms 5 z 5 z2 Q the deﬁning equations j 1 j j 1 j 0of 0 into the single equation

3 ξ1ξ4 + ξ2ξ3 0, [ξ1 : ··· : ξ4] ∈ CP . (The Lagrange substitution will be given explicitly in Subsection G; cf. also the analogous case in Section 6 for solving cubic equations.) This equation deﬁnes the complex surface

3 Q {[ξ1 : ··· : ξ4] ∈ CP | ξ1ξ4 + ξ2ξ3 0}

3 in CP . In what follows, we will identify Q0 with Q under this linear equivalence. For each value of a parameter c∗ ∈ Cˆ , the equations ξ ξ − 1 3 c∗ ξ2 ξ4 deﬁne a complex projective line in Q; we call this a generating line of the ﬁrst kind (with parameter c∗). In a similar vein, for c∗∗ ∈ Cˆ , This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

366 25. The Icosahedron and the Unsolvable Quintic

the equations ξ ξ 1 − 2 c∗∗ ξ3 ξ4 define in Q a generating line of the second kind (with parameter c∗∗). As can be seen from the defining formulas, the two families of generating lines satisfy the following properties: (1) Each point of the quadric is the intersection of two generating lines of different kinds; (2) any two generating lines of different kinds intersect at exactly one point; (3) any two distinct generating lines are disjoint. (For a (more than) good analogy, consider a hyperboloid of one sheet as a doubly ruled surface with the two rulers corresponding to the generating lines of the first and second kind.) Due to the linear equivalence between Q and Q0, the entire construction in Q can be carried over to our initial quadric Q0. ∗ ∗∗ The parameters c and c become quotients of linear froms in the z ,...,z 5 z 5 z2 variables 1 5 subject to j 1 j j 1 j 0. If we fix a point o ∈ Q0 as the origin, then the generating lines CP∗ and CP∗∗ of the first and second kind through o can be viewed as axes of a “rectilinear” coordinate system for Q0. With respect to this coordinate system, any point in Q0 can be uniquely represented by a pair of complex coordinates (c∗,c∗∗) ∈ CP∗ × CP∗∗ in an obvious manner. This gives a conformal equivalence ∗ ∗∗ Q0 CP × CP .

4 Recall that the symmetric group S5 acts on CP by permuting the homogeneous coordinates. In view of the symmetries, this ac- Q ⊂ P3 S tion leaves 0 C 0 invariant. Each permutation in 5 maps generating lines to generating lines. This follows from well-known projective geometric considerations, or by the explicit formulas 4 given in Subsection G. In fact, the elements of S5 act on CP as projective transformations, since they act on C5 as permutation matrices. In particular, restricted to Q0, the elements of S5 act as projective collineations; i.e., they take projective lines to projective lines. By continuity, each collineation in S5 either maps the generating lines within a family to generating lines in the same family, or interchanges the generating lines between the two fam- This spread is one line short. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

F. Geometry of the Canonical Equation: General Considerations 367 ilies. Let G ⊂ S5 be the subgroup that preserves the generating lines in each family. We claim that G A5. It is clear that the index of G in S5 is at most 2. Since the alternating group A5 is the only index 2 subgroup in S5, it follows that A5 ⊂ G. For the reverse inclusion, notice that G acts on CP∗ by complex automorphisms. This realizes G as a subgroup of Aut (CP∗), where the latter is the group of all complex automorphisms of the projective line CP∗. The choice of an inhomogeneous coordinate in CP∗ identifies CP∗ with Cˆ , and Aut (CP∗) with the Mobius¨ group Mob¨ (Cˆ ). This identification makes G a finite subgroup of Mob¨ (Cˆ ). On the other hand, ˆ by Theorem 18, the largest finite subgroup of Mob¨ (C) is A5. Thus G A5 follows. In particular, we also obtain that the collineations that correspond to the odd permutations in S5 do interchange the two families of generating lines. By the very definition of the conformal equivalence above, the action of A5 on the two families of generating lines induces an ∗ ∗∗ action of A5 on both CP and CP such that the conformal equivalence is A5-equivariant, where A5 is considered to act on the product CP∗ × CP∗∗ diagonally. As we saw above, the action of A5 on generating lines of the first ∗ and second kind realizes A5 as a subgroup in Aut (CP ) and also in ∗∗ Aut (CP ). In a similar vein, an odd permutation in S5 gives rise to ∗ ∗∗ a conformal equivalence of CP and CP . Conjugating A5 by this ∗∗ odd permutation (within S5) then carries the action of A5 on CP ∗ into an action of A5 on CP , and this latter action is equivalent to ∗ the original action of A5 on CP via complex automorphisms of CP∗. ∗ ∗ ∗∗ We now consider the projections π : Q0 → CP and π : Q0 → CP∗∗. Let    5 5  Q˜ (z ,...,z ) ∈ 5 −{ }| z z2 , 0  1 5 C 0 j j 0 j 1 j 1

˜ and deﬁne π : Q0 → Q0 to be the restriction of the canonical projection C5 −{0}→CP4.Weset

z∗ π∗ ◦ π and z∗∗ π∗∗ ◦ π. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

368 25. The Icosahedron and the Unsolvable Quintic

We take a closer look at z∗. An inhomogeneous coordinate on CP∗ identiﬁes CP∗ with Cˆ , and z∗ can be viewed as the composition

π π∗ 5 ˜ ∗ ˆ (C −{0}⊃) Q0→Q0→CP C(⊃ C). Q Q z∗ In view of the linear equivalence of 0 and , we see that is z ,...,z 5 z a rational function in the variables 1 5 subject to j 1 j 5 z2 A j 1 j 0. By construction, 5 acts on these variables by even ∗ permutations, and this induces an action of A5 on CP by complex automorphisms. With a choice of an inhomogeneous coordinate on CP∗, this latter action is by linear fractional transformations. This ˆ identifies A5 with a subgroup of Mob¨ (C). Different choices of inhomogeneous coordinates on CP∗ give rise to conjugate subgroups in Mob¨ (Cˆ ). By Theorem 18, there is an inhomogeneous coordinate ∗ on CP with respect to which A5 is identified with the icosahedral Mobius¨ group I. From now on we assume that this choice has been made, and we let ϕ : A5 → I denote the corresponding isomor- ∗ ˜ ˆ phism. Summarizing, we see that z : Q0 → C is ϕ-equivariant, ˜ where A5 acts on Q0 by permuting the coordinates, and I acts on Cˆ as the icosahedral Mobius¨ group. When K is considered as the splitting field of a canonical quintic 5 2 P(z) z + az + bz + c with roots z1,...,z5 and Galois group A5, ∗ then z (z1,...,z5) becomes an element of K k(z1,...,z5), since ∗ ∗ z depends rationally on z1,...,z5. Since z is ϕ-equivariant, the 60 roots of the resolvent polynomial are nothing but the icosa- ∗ hedral linear fractional tranformations applied to z (z1,...,z5). Summarizing, we have shown that z∗ satisfies the icosahedral equation √ ∗ ∗ q(z (z1,...,z5)) w (a,b,c, δ),

z ,...,z 5 z 5 z2 where 1 5 are subject to j 1 j j 1 √j , and the parame- ∗ ter w on the√ right-hand side depends on a, b, c, δ rationally. (The presence of δ is due to the fact that the Galois group is A5 not S5.) To complete the proof of Theorem 19, we need to show that z∗ generates K over k. (Along with the claimed technical details, we will do this in the next subsection.) The situation is completely analogous for the generating lines of the second kind. We obtain an Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

G. Geometry of the Canonical Equation: Explicit Formulas 369

∗∗ element z (z1,...,z5) ∈ K that satisﬁes the icosahedral equation √ ∗∗ ∗∗ q(z (z1,...,z5)) w (a,b,c, δ).

The parameters√a, b, c are invariant under the entire symmetric group S5, while δ changes its sign when the roots are subjected ∗ ∗∗ to odd permutations. Since the two actions of A5 on CP and CP are conjugate under the odd permutations in S5, we obtain √ √ w∗(a,b,c,− δ) w∗∗(a,b,c, δ).

G. Geometry of the Canonical Equation: Explicit Formulas

To exhibit the stated linear equivalence between Q0 and Q,weﬁrst deﬁne ι : CP3 → CP4 by

ι([ξ1 : ··· : ξ4]) [z1 : ··· : z5], where

4 −(j−1)l zj ω ξl,j 1,...,5. l 1

Since 1 + ω + ω2 + ω3 + ω4 0, we have 5 4 5 −(j−1)l zj ω ξl 0. j 1 l 1 j 1

ι P3 P3 ⊂ P4 Thus, the linear map sends C into the linear slice C 0 C . ι P3 P3 Actually, is a linear isomorphism between C and C 0 . As simple ι−1 P3 → P3 computation shows, the inverse : C 0 C is given by

5 1 (j−1)l ξl ω zj,l 1,...,4. 5 j 1 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

370 25. The Icosahedron and the Unsolvable Quintic

5 z2 Q To translate the deﬁning equation j 1 j 0of 0 in terms of the ξl’s, we compute 5 5 4 2 2 −(j−1)l zj ω ξl j 1 j 1 l 1 4 5 −(j−1)(l+l) ω ξlξl l,l 1 j 1

10(ξ1ξ4 + ξ2ξ3).

5 ω−(j−1)(l+l) l + l The last equality holds because j 1 5 iff 5, and zero otherwise. This shows that the quadrics Q and Q0 correspond to each other. Given that the variables ξl, l 1,...,4, are linear forms in z1,z2,z3,z4,z5, the equations for the generating lines give explicit ∗ ∗∗ rational dependence of z and z on z1,z2,z3,z4,z5. Due to our explicit formulas, we can determine the linear fractional transfor- ∗ ∗∗ mations that z and z undergo when the variables z1,z2,z3,z4,z5 are subjected to even permutations. It is enough to work this out for π∗, since the homogeneous and inhomogeneous coordinates are permuted in the same way. (The case of π∗∗ can be treated analogously.) We will give explicit formulas only for the generators S and W of the icosahedral group. We claim that S (multiplication by ω) corresponds to the cyclic permutation

S : zj → zj( mod 5)+1,j 1,...,5. Indeed, applying the latter, we have

5 j−1 5 (j−1) ξ ω zj ω zj( )+ c∗ − 1 − j 1 →− j 1 mod 5 1 ξ 5 ω2(j−1)z 5 ω2(j−1)z 2 j 1 j j 1 j( mod 5)+1

5 j−2 ω zj − j 1 ωc∗, 5 ω2(j−2)z j 1 j and the claim follows. In a similar vein, W corresponds to the permutation

W : z1 → z1,z2 ↔ z3,z4 ↔ z5. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

G. Geometry of the Canonical Equation: Explicit Formulas 371

Since this is tedious, we give a proof. Using the explicit form of W derived in Section 18, and Problem 5(b) of Section 17, we have (ω − ω4)ξ + (ω2 − ω3)ξ τξ + ξ Wc∗ − 1 2 − 1 2 , 2 3 4 (ω − ω )ξ1 − (ω − ω )ξ2 ξ1 − τξ2 where τ is the golden section. On the other hand, permuting the zj’s according to the recipe above, and rewriting the corresponding quotient in terms of the ξl’s, we have z + ω2z + ωz + ω4z + ω3z − 1 2 3 4 5 4 2 3 z1 + ω z2 + ω z3 + ω z4 + ωz5 ( + ω + ω4)ξ + ( + ω2 + ω3)ξ + ( + ω + ω4)ξ + ( + ω2 + ω3)ξ − 1 2 2 1 3 2 3 3 1 2 2 4 2 3 2 3 4 4 (3 + ω + ω )ξ1 + (1 + 2ω + 2ω )ξ2 + (1 + 2ω + 2ω )ξ3 + (3 + ω + ω )ξ4 ( + /τ)ξ + ( − τ)ξ + ( + /τ)ξ + ( − τ)ξ − 1 2 1 3 2 3 1 3 1 2 4 (3 − τ)ξ1 + (1 − 2τ)ξ2 + (1 + 2/τ)ξ3 + (3 + 1/τ)ξ4 ξ + ξ /τ + ξ τ − ξ − 1 2 3 4 ξ1/τ − ξ2 + ξ3 + ξ4τ (τξ + ξ )( /τ + ξ /ξ ) τξ + ξ − 1 2 1 3 1 − 1 2 . (ξ1 − τξ2)(1/τ + ξ3/ξ1) ξ1 − τξ2 √ √ √ Here we used√ 1 + 2/τ 5, 3 − τ 5/τ,3+ 1/τ 5τ, 1 − 2τ − 5, and ξ1ξ4 + ξ2ξ3 0. The permutation rule for W follows. Comparing how S and W transform the zj’s and the tj’s, we see that when z∗ is subjected to the linear fractional transformations of the icosahedral group I, then this action can be realized as even permutations on its variables in exactly the same manner as A5 acts on the roots t1,...,t5 of the icosahedral resolvent. This means that an A5-equivariant one-to-one correspondence zj ↔ tj, j 1,...,5, can be established between the two sets {z1,...,z5} and {t1,...,t5}. j (z1 and t1 are the unique ﬁxed points of W, and zj+1 S (z1) j corresponds to tj+1 S (t1), j 1,...,4.)

Remark. The odd permutation

z1 → z1,z2 → z4,z3 → z2,z4 → z5,z5 → z3 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

372 25. The Icosahedron and the Unsolvable Quintic

has the effect of replacing ω by ω2 and interchanging c∗ and c∗∗. Thus, the formulas for the generating lines of the second kind can be derived from those of the ﬁrst kind by the substitution ω → ω2.

Remark. The Lagrange substitutions above that establish the linear equivalence between the quadrics Q and Q0, as well as make the structure of the generating lines transparent, are derived in the Icosahedron Book based on the requirement that the transformation rules for S and W should match with those of the roots of the quintic icosahedral resolvent. Here we followed a somewhat ad hoc but quicker approach.

To get a closer look at the correspondence above, we now work out the generating lines in terms of the roots {t1,...,t5} of the icosahedral resolvent. In perfect analogy with the Lagrange substitutions, we put

4 −(j−1)l tj ω Xl,j 1,...,5, l 1

and

5 1 (j−1)l Xl ω tj,l 1,...,4. 5 j 1

To work out Xl, we use the explicit forms of Vj and TjVj derived in Subsection C. Substituting them into the expression for tj above, we obtain

4(j−1) 3(j−1) 2(j−1) j−1 tj (ω z1 − ω z2)A + (ω z1 + ω z2)B,

where A, B are linear in u and v. (Here z1 and z2 are the complex arguments of our forms with z z1/z2. Do not confuse them with the ﬁrst two roots of the canonical quintic! In any case, z1 and z2 appear here only briy.) With this, the inverse of the Lagrange Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

G. Geometry of the Canonical Equation: Explicit Formulas 373 substitution becomes 5 1 (j−1)l 4(j−1) 3(j−1) Xl ω (ω z1 − ω z2)A 5 j 1

5 1 (j−1)l 2(j−1) j−1 + ω (ω z1 − ω z2)B 5 j 1

(δ1lz1 − δ2lz2)A + (δ3lz1 + δ4lz2)B, where we used the Kronecker delta function δjl ( 1 iff j l and zero otherwise). Writing the cases out, we have

X1 z1A, X2 −z2A, X3 z1B,X4 z2B. For the parameters C∗ and C∗∗ of the generating lines deﬁned by X X X X − 1 3 C∗, 1 − 2 C∗∗, X2 X4 X3 X4 we obtain z A C∗ 1 z, C∗∗ . z2 B Since z∗ and z∗∗ are essentially given by the projections (C∗,C∗∗) → C∗ and (C∗,C∗∗) → C∗∗, we thus have A z∗(t ,...,t ) z z∗∗(t ,...,t ) . 1 5 and 1 5 B The ﬁrst equation is of paramount importance. Writing the canonical resolvent in the short form 5 ∗ 5 2 P (X) (X − tj) X + 5a(u, v, w)X + 5b(u, v, w)X + c(u, v, w), j 1 we have ∗ ∗ q(z (t1,...,t5)) w (a(u,v,w),b(u,v,w),c(u,v,w), δ(u, v, w)) q(z) w.

Here δ(u, v, w) is the discriminant of the canonical quintic resolvent. (It has the same expression as δ in Subsection C with a, b, c replaced by a(u, v, w), b(u, v, w), c(u, v, w).) Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

374 25. The Icosahedron and the Unsolvable Quintic

The system a(u, v, w) a, b(u, v, w) b, c(u, v, w) c,

can be inverted, provided that we√ also match the square roots of the discriminants δ(u, v, w) δ: √ u u(a, b, c, δ), √ v v(a, b, c, δ), √ w w(a, b, c, δ).

We will do this inversion explicitly at the end of this subsection. By solving this system we attain that the roots of the canonical quintic P and the roots of the quintic resolvent P∗ coincide as sets. On the other hand, we saw that there exists an A5-equivariant correspondence between these sets of roots. Imposing this we obtain

zj tj,j 1,...,5. With this we can now describe how to solve a given irreducible quintic. First we use the Tschirhaus transformation to reduce the quintic to a canonical form P(z) z5 +5az2 +5bz +c. This amounts to solving a quadratic equation. We also compute the discriminant δ from the explicit form given in Subsection C. Then we substitute the coefﬁcients a, b, c into the right-hand sides of the equations in the inverted system above and obtain u, v, and w. We now solve the icosahedral equation for this particular value of w to obtain z as a ratio of hypergeometric functions as in Subsection B. By working out the forms Tj and Vj using the particular values of u and v and z we obtain

tj tj(u, v, z), j 1,...,5.

Since tj zj, these are the ﬁve roots of our quintic. Tracing our ∗ steps back, we see that z1,...,z5 depend rationally on z . In particular, when K k(z1,...,z5) is the splitting ﬁeld of the canonical Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

G. Geometry of the Canonical Equation: Explicit Formulas 375 quintic, we obtain that z∗ generates K over k. This was the missing piece in the proof of Theorem 19. Our final task is to invert the system above. Using the explicit forms of the coefficients of the canonical quintic in Subsection D, we write the system as 6uv2 + v3 w · a 8u3 + 12u2v + , 1 − w w · b 6u2v2 + 4uv3 3v4 −4u4 + + , 3 1 − w 4(1 − w)2 w · c 40u3v2 15uv4 + 4v5 48u5 − + . 3 1 − w (1 − w)2 A small miracle (due to our ad hoc approach) happens here. In trying to cancel the top terms in u and v, if we multiply the first equation by −4v2/(1 − w) and the second by 12u, and then add all three then everything cancels on the right-hand sides! Rearranging the terms on the left-hand side, we obtain v2 ub + c 12 . 1 − w 12a We call this the fundamental relation. Continuing, we can create cubic binomial terms as follows: v2 9 v2 3 −uc + b − 4u2 − 1 − w 4w 1 − w and 4 1 − w 1 v2 3 a2 − (3ua + 2b)2 4u2 − . 81 v2 w 1 − w Combining these, we obtain 4 1 − w 4 v2 a2 − (3ua + 2b)2 uc − b . 81 v2 9 1 − w Using the fundamental relation, this becomes 16 a 4 12ub + c a2 − (3ua + 2b)2 uc − b . 27 12ub + c 9 12a Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

376 25. The Icosahedron and the Unsolvable Quintic

This is a quadratic equation in u! The quadratic formula gives √ √ √ ( a3b + b2c − ac2) ± a δ/( ) u u(a, b, c, δ) 11 2 25 5 , 24(a4 − b3 + abc) √ where the sign in front of δ needs to be determined.√ Since u is now known in terms of a, b, c, and δ, we can use the fundamental relation in the first equation containing a cubic binomial term and solve for w. We obtain √ ( u2a − ub − c)2 w w(a, b, c, δ) 48 12 . 64a2(12u(ac − b2) − bc) Finally, the first equation of our system can be rewritten as v2 v2 12u2 + v wa − 8u3 − 6u . 1 − w 1 − w Using once again the fundamental relation, we finally arrive at √ u3a + u2b + uc − a2w v v(a, b, c, δ) −96 72 6 12 . 144u2a + 12ub + c √ √ To determine the sign of δ,weset δ δ(u, v, w), where 2 δ(u, v, w) (tj − tl) 1≤j

is the discriminant of the canonical resolvent. It is enough to work out δ(u, v, w) for u 1 and v 0. For these values of the parameters our system reduces to

a 8 ,b −12 ,c 144 . w w w Substituting these and u 1 into the formula for u with the sign ambiguity, we obtain √ √ 1 − w ± δ 124 · 25 5 . w3

Remark. Note that as a good check of our computations we can also substitute the values of a, b, c above into the discriminant formula in Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 377

Subsection C and get (1 − w)2 δ 128 · 55 . w6

On the other hand, for u 1 and v 0, tj reduces to sj 12 (I/H)Vj, and hence, I10 10 δ(1, 0,w) (sj − sl) 12 (Vj − Vl). H10 1≤j

Problems

1. Verify the following identities: √ Hess (R) −48 3iS, √ Hess (S) 48 3iR, √ Jac (R, S) 32 3iT, Hess (T) −25RS, Jac (T, R) −4S2, Jac (T, S) −4R2.

2. Interpret the zeros of the degree-8 forms Vj, j 1,...,5, geometrically. 3. Locate the axes of the half-turns SjUS−j, SjVS−j, SjWS−j, j 1,...,5, in our model of the icosahedron. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

378 25. The Icosahedron and the Unsolvable Quintic

4. (a) Work out the resolvent of the ﬁve octahedral forms using the following steps. First consider the product 5 5 4 3 2 (X − Tj) X + a1X + a2X + a3X + a4X + a5, j 1 where X is used as a variable. Using the description of the invariants of the icosahedral group, conclude that this quintic resolvent is X 5 − 10IX3 + 45I2 X − J . Introduce the new variable I2 X 12 X J

(depending only on z z1/z2), and use the second table in Section 24 to arrive at the quintic icosahedral resolvent polynomial 5 P∗(X) (X − r∗) X5 − 5 X3 + 5 X − 1 . j ( − w) ( − w)2 ( − w)2 j 1 6 1 16 1 12 1 (r∗(w),...,r∗(w)) ∈ 5 (b) Describe the geometry of the “solution set” 1 5 C as w varies in Cˆ and q(z) w as follows. Introduce the sums of various powers

5 ∗ l σl (rj ) ,l 1,...,4, j 1 and verify 5 5 σ 0,σ ,σ 0,σ . 1 2 3(1 − w) 3 4 36(1 − w)2 2 Notice that σ 20σ4. Deﬁne 2    5 5  D r∗ ... r∗ ∈ P4 | r∗ (r∗)3 [ 1 : : 5 ] C j j 0 j 1 j 1 and    5 5 2 5  F r∗ ··· r∗ ∈ P4 | r∗ , (r∗)2 (r∗)4 , [ 1 : : 5 ] C j 0 j 20 j  j 1 j 1 j 1

and prove that D and F are smooth algebraic surfaces in CP3 of degree 3 and P3 P3 P4 4, respectively, (C is identified with the linear slice C 0 of C defined by 5 r∗ j 1 j 0). Show that the projective points that correspond to the solutions of the quintic icosahedral resolvent above fill the smooth algebraic curve D ∩ F ⊂ CP3 of degree 12. (This problem is rich in history. The quintic resolvent for X is called the Brioschi quintic. The surface D was named by Clebsch the diagonal surface, since this surface contains 15 diagonals of the coordinate pentahedron. The intersection of the diagonal surface D with the canonical Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Problems 379

surface Q is a (smooth) curve called the Bring–Jerrard curve.6 A generating line through the point [z1 : ··· : z5] ∈ Q, corresponding to the roots of a canonical quintic, intersects the Bring–Jerrard curve in 3 points, and these intersection points can be obtained by extracting suitable square and cube roots. There is a beautiful geometric interpretation of the Bring–Jerrard curve in terms of stellated dodecahedra7.) 5. (a) Derive an icosahedral resolvent polynomial of degree 6 based on the 6 quartic forms that vanish on the 6 pairs of antipodal vertices of the icosahedron φ z2z2 Sj φ as follows. Set ∞ 5 1 2 , and apply the icosahedral substitutions to ∞ to obtain φ (z ,z ) (ωjz2 + z z − ω4jz2)2,j ,..., . j 1 2 1 1 2 2 0 4 Comparing coefﬁcients, show that the resolvent form satisﬁes φ6 − 10Iφ3 + Hφ + 5I2 0.

(b) Work out the action of the icosahedral generators S and W on φ∞ and φj, j 0,...,4, and verify

S : φ∞ → φ∞,φj → φj+1( mod 5),j 0,...,4;

W : φ∞ ↔ φ0,φ1 ↔ φ4,φ2 ↔ φ3. Interpret these transformation rules as congruences S : j ≡ j + 1( mod 5),

W j ≡−1 ( ). : j mod 5 Show that the icosahedral group (the Galois group of this resolvent) is isomorphic to SL(2, Z5)/{±I}: aj + b j ≡− ( ), ad − bc ≡ ( ), a, b, c, d ∈ . cj + d mod 5 1 mod 5 Z

6For a very detailed exposition, see R. Fricke, Lehrbuch der Algebra, Vol. II, Braunschweig, 1926. For a recent text, see J. Shurman, Geometry of the Quintic, John Wiley and Sons, 1997. 7See P. Slodowy, Das Ikosaeder und die Gleichungen f¨unften Grades, in Mathematische Miniaturen,¨ Band 3, Birkhauser-Basel,¨ 1986. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... The Fourth . . . . . Dimension . 26 . SECTION .

♣ In Section 22, we encountered a number of objects in the 4- dimensional Euclidean space, such as S3. Although it sits in R4, S3 is essentially a 3-dimensional object. In fact, the generalized 3 stereographic projection gives rise to a diffeomorphism hN : S − {N}→R3 where N (0, 0, 0, 1) is the North Pole and R3 is the linear subspace of R4 spanned by the ﬁrst three coordinate axes. (In coordinates, the inclusion R3 ⊂ R4 is given by assigning to 3 4 (a,b,c) ∈ R the point (a,b,c,0) ∈ R .) The deﬁnition of hN is the 3 same as its 2-dimensional brother; hN (p), N p ∈ S , is the unique intersection point of R3 and the line through N and p. (Notice that N, O, p and hN (p) all lie in a 2-dimensional plane which cuts a great 3 4 circle from S ⊂ R .) The stereographic projection hN enables us to visualize objects that lie in S3 as objects in R3; in fact, this was used to create Color Plate 10 (Clifford tori).

Remark. ♦ Quaternions can be used to give a simple and explicit form of the stereographic projection. If we let R3 correspond to the linear space H0 of pure quaternions and 1 ∈ H correspond to the North

380 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

26. The Fourth Dimension 381

3 Pole N, then hN : S → H0 is given by q −!(q) 3 hN (q) , 1 q ∈ S . 1 −!(q)

Indeed, since 1, q, and hN (q) are on the same line, we have

t(1 − q) 1 − hN (q) for some t>0. Taking the real part of both sides and using the fact that hN (q) is pure, we obtain

t 1 . 1 −!(q) Substituting this back, the formula follows.

♣ The purpose of this section is to give a few examples that will enable us to get a feel for objects whose natural environment is R4. We begin with a cube1 in R4. The best way to understand how a 4-dimensional cube and an ordinary 3-dimensional cube are related is to observe how a 3-dimensional cube is derived from a 2- dimensional cube, i.e., a square. As for the latter, given a ﬂat square in space, choose a vector perpendicular to the plane of the square (Figure 26.1). We assume that the magnitude of the vector is given by the side length of the square. Performing spatial translation with this vector, we see that the square sweeps a 3-dimensional cube. During this process the number of vertices doubles, the number of edges doubles plus the old vertices sweep four new vertices, and ﬁnally, the number of faces doubles plus the old edges sweep four new faces. Obtaining a 4-dimensional cube is now remarkably simple; just repeat this process! Take a 3-dimensional cube in R4, choose a vector perpendicular to the 3-dimensional linear subspace spanned by the cube and watch the object that is swept by translating the cube along this vector (Figure 26.2). In particular the same counting principles hold, and we see that a 4-dimensional cube has 2 × 8 vertices, (2 × 12) + 8 edges, and (2 × 6) + 12 faces!

1The ﬁrst pioneers of 4-dimensional geometric vision were Mobius¨ (c. 1827), Schli,¨ Cayley, and Grassmann (c. 1853). Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

382 26. The Fourth Dimension

Figure 26.1

The skeptical reader may now say, “This is all humbug; the 4- dimensional cube is drawn on a 2-dimensional sheet of paper, so there is no way to get a feel for this object!” It is true, the picture you see is the projection of a 4-dimensional cube to a 2-dimensional plane, but remember that we use the same trickery every day on the chalkboard to visualize 3-dimensional objects!

Remark. Observing how the 3-dimensional tetrahedron is derived from the 2-dimensional triangle, it is easy to obtain a model of the 4- dimensional tetrahedron. We already encountered this object in Section 20; in fact, the pentagonal graph in Figure 20.4 gives a 2- dimensional projection of the regular tetrahedron in R4. (Can you identify the ﬁve tetrahedral faces?)

Figure 26.2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

26. The Fourth Dimension 383

All 4-dimensional regular polytopes2 were discovered and classified by Schli¨ around 1853. Bypassing some computational details, we now sketch3 this classification. Note first that the relevant concepts (flag, regularity, vertex figure, reciprocality, etc.) introduced in Section 17 for regular polyhedra can be extended to convex polytopes in 4 dimensions. In particular, a (4-dimensional) polytope has vertices, edges, faces, and 3-dimensional faces, called cells. For example, the 4- dimensional tetrahedron, or pentatope, has 1 + 4 tetrahedral cells, and the 4-dimensional cube, or hypercube, has 2 + 6 cubical cells. The Schli¨ symbol of a regular polytope is a triple {a, b, c}, where {a, b} is the Schli¨ symbol of a typical cell, and each edge of the polytope is shared by c cells. A closer look at Figure 20.4 and Figure 26.2 shows that the Schli¨ symbol of the pentatope is {3, 3, 3}, and the Schli¨ symbol of the hypercube is {4, 3, 3}. By a bit of a stretch of Euclid’s argument in Section 17 we can list all possible Schli¨ symbols. Take c congruent cells (regular polyhedra) of Schli¨ symbol {a, b}, and cluster them around a single common edge (with consecutive cells pasted together along a single commmon face) in a fanlike pattern. This can be done in 3 dimensions. (For analogy, paste three, four, or five congruent equilateral triangles together to form a fanlike pattern sharing a single common vertex in the plane.) There are exactly two faces (one for the initial cell, and one for the terminal cell) in the fan sharing the common edges that have not been pasted together. Now place the entire configuration in 4 dimensions, and paste these remaining two faces together. Notice that as you move the two faces together, the entire configuration pops out in the fourth dimension! (Con- tinuing the analogy, the fans of the equilateral triangles above pop out in 3 dimensions to form the sides of triangular, square, and pentagonal pyramids, the basic ingredients to make the tetrahedron, octahedron, and icosahedron!) For the existence of the fan of c cells, it is clearly necessary that cδ < 2π,

2In 4-dimensions we use the word polytope rather than polyhedron. 3For a detailed account, see H.S.M. Coxeter, Regular Polytopes, Pitman, 1947. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

384 26. The Fourth Dimension

where δ is the dihedral angle of the cell with Schli¨ sybol {a, b}. As we will see below, this constraint will give all possible Schli¨ symbols {a, b, c} of a regular polytope. Thus, we need to work out the dihedral angle for each of the Platonic solids. We begin with the tetrahedron. In Figure 17.16, the grey triangle (cut out from the tetrahedron by a symmetry plane) is isosceles, and its larger angle at the midpoint√ of the edge is the dihedral angle δ. We have sin(δ/2) 1/ 3, or δ ≈ 70.5287793◦. The constraint above gives 3, 4, 5 as the possible values of c. We obtain that the possible Schli¨ symbols for a regular polytope with tetrahedral cells are {3, 3, 3}, {3, 3, 4}, and {3, 3, 5}. The dihedral angle of a cube is δ π/2, so that c 3 and the only possible Schli¨ symbol is {4, 3, 3}. A symmetry plane of the octahedron containing two opposite vertices and two midpoints of sides cuts out a rhombus from the octahedron, and δ is the larger√ angle of the rhombus at the midpoints. We have tan(δ/2) 2, or δ ≈ 109.4712206◦. Thus, c 3, and the only possible Schli¨ symbol is {3, 4, 3}. A good look at Figure 17.30 shows that the dihedral angle of the icosahedron satisfies tan(δ/2) τ/(τ − 1), where τ is the golden section. A bit of computation gives δ ≈ 138.1896851◦. We see that there cannot be a regular polytope with icosahedral cells. Finally, the dihedral angle of the dodecahedron can be obtained from the roof-proof (Problem 20 of Section 17). The perpendicular bisector of the ridge of a roof is a symmetry plane that cuts out an isosceles triangle from the roof, and the larger angle (at the midpoint of the ridge) of this triangle is δ. From the metric properties of the roof, we have tan(δ/2) τ,orδ ≈ 116.5650512◦. We obtain c 3 and the Schli¨ symbol {5, 3, 3}. Summarizing, we see that the possible Schli¨ symbols of a regular polytope are {3, 3, 3}, {4, 3, 3}, {3, 3, 4}, {3, 4, 3}, {5, 3, 3}, {3, 3, 5}. The first two give the Schli¨ symbols of the pentatope and the hypercube. Amazingly, the rest are also realized as Schli¨ symbols of regular polytopes! To complete the classification, we now give brief geometric descriptions for each of these new pentatopes. (For pictures, see H.S.M. Coxeter, Regular Polytopes, Pitman, 1947, or Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

26. The Fourth Dimension 385

D. Hilbert and S. Cohn-Vossen, Geometry and Imagination, Chelsea, New York, 1952.) Recall that 4 alternate vertices of the cube are the vertices of an inscribed tetrahedron. Each omitted vertex corresponds to a face of the tetrahedron. (For an omitted vertex, the vertex figure of the cube is parallel to the face.) Analogously, 8 alternate vertices of a hypercube are the vertices of a regular polytope. This polytope has 16 tetrahedral cells, and for this reason it is called the 16-cell. Eight tetrahedra correspond to each of the omitted vertices, and 8 other tetrahedra are inscribed in each of the 8 cubic cells of the hypercube. The 16-cell has 24 edges that are the diagonals of the 24 square faces of the hypercube. (Since we are selecting alternate vertices of the hypercube, we get only one diagonal for each square face.) It follows that each edge must be surrounded by 4 tetrahedra, so that the Schli¨ symbol is {3, 3, 4}. The 16-cell has 16 × 4/2 32 triangular faces. A plane projection of the 16-cell can be obtained by starting with the vertices of a regular octagon, and connecting all but opposite pairs of vertices by line segments. The regular polytope with Schli¨ symbol {3, 4, 3}, the so-called 24-cell, can be obtained from the 16-cell as follows. The vertices of the 24-cell are the midpoints of the 24 edges of the 16-cell. It has 24 octahedral cells. Eight of these cells are the vertex figures of the 16-cell, and 16 are inscribed in the 16 tetrahedra of the 16-cell. (For analogy, in Figure 17.20 of the truncated tetrahedron, 4 of the 8 faces of the inscribed octahedron are the vertex figures of the tetrahedron, and the 4 remaining faces are inscribed in the faces of the tetrahedron.) Each edge of the 24-cell is surrounded by 3 octahedra. The 24-cell has 24 × 12/3 96 edges and 24 × 8/2 96 triangular faces. It is easier to visualize the remaining two regular polytopes projected to the 3-sphere S3 from the origin. For the spherical regular polytope with Schli¨ symbol {5, 3, 3}, we first construct a typical cell, a spherical dodecahedron in S3. Recall from above that the dihedral angle of the Euclidean dodecahedron is δ ≈ 116.5650512◦.If this dodecahedron is small, its metric properties are close to those of the spherical dodecahedron of the same size. At the other extreme, if a small spherical dodecahedron is iated, at one stage it Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

386 26. The Fourth Dimension

will cover an entire hemisphere in S3, and hence the dihedral angle will become 180◦. By continuity, there must be a spherical dodecahedron whose dihedral angle is 120◦. As expected, this puffy (and tiny) dodecahedron is almost indistinguishable from its Euclidean brother. Due to the fact that the dihedral angle is 120◦ 2π/3, exactly 3 of these dodecahedra fit together to share a common edge. Hence, if we are able to tesselate S3 with spherical dodecahedra of this size, the regular polytope obtained must have Schli¨ symbol {5, 3, 3}. It turns out that this construction is possible, and we need 120 dodecahedra for the tesselation. For this reason, this polytope is called the 120-cell. The explicit construction of the 120-cell is technical. There is an easy way, however, to see how these 120 dodecahedra fit together in S3, and it is based on the fact that, up to adjustment by an isometry, the centroids of the dodecahedral cells can be considered as the 120 elements of the binary icosahedral group I∗ in S3 discussed in Section 23. Recall that in terms of the Clifford decomposition of S3, I∗ is made up of the vertices of two regular decagons inscribed in the orthogonal circles C±1, and the rest appear (in two C √ groups of 50) in the Clifford tori ±1/ 5 (Figure 23.5). In view of this, the 120-cell can be constructed as follows. First make a “necklace” of 10 (spherical) dodecahedra such that the centroids of the ∗ dodecahedra are the 10 elements of I on C1. It turns out that these dodecahedra have dihedral angle 120◦ as above. A pair of consecutive dodecahedra in the necklace are pasted together at a common pentagonal face. Each of the 5 edges of this common face is the shared edge of two other pentagonal faces, one from each of the consecutive dodecahedra. These two faces meet at a dihedral angle of 120◦, so that another dodecahedron can be pasted in. Since we have five edges (of the common pentagonal face), we can paste in 5 extra dodecahedra around the two consecutive dodecahedra in the necklace. This cluster of 5 dodecahedra makes a “bulge” in the necklace. Since the necklace has 10 places (of consecutive dodecahedra) for this construction, we can add 10 bulges to the necklace, a total of 10 × 5 dodecahedra. These, along with the original 10 dodecahedra, use up 60 dodecahedra, and give a “bumpy” polyhe- Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

26. The Fourth Dimension 387 dral Clifford torus in S3. As computation shows, the centroids of the dodecahedra in the bulges make up the 50 elements of I∗ in C √ 1/ 5. Finally, the entire construction can be repeated for C−1 and C √ −1/ 5, and the two bumpy Clifford tori fit together to form the 120-cell. (This visualization of the 120-cell also reveals that the faces of each dodecahedral cell are contained in the perpendicular bisectors of the line segments connecting the centroid (in I∗)ofthe cell and the 2+5+5 nearby elements in I∗.) The 120-cell has 120 × 12/2 720 faces (by double counting the faces of the 120 dodecahedra), 120 × 30/3 1200 edges (since 3 dodecahedra share a common edge), and 120×20/4 600 vertices (since 4 dodecahedra meet at a common vertex). The reciprocal of the 120-cell is the 600-cell, a regular polytope with 600 tetrahedral cells and Schli¨ symbol {3, 3, 5}. A typical cell is a puffy spherical tetrahedron with dihedral angle 72◦.Asthe Schli¨ symbol suggests, exactly 5 share a common edge. At each vertex exactly 20 tetrahedra meet. (Indeed, consider the configuration of rays tangent to the edges at a vertex. This configuration intersected with a sphere around the vertex (within the tangent space of S3 at the vertex) gives the vertices of a regular polyhedron. This polyhedron must have the largest number of vertices; that is, it must be a dodecahedron.) Since each of the 600 tetrahedra contributes 4 vertices to the tesselation, and this way the total number of vertices is overcounted 20 times, we see that the number of vertices of the tesselation is 600 × 4/20 120. Once again, it turns out that up to an adjustment of the tessellation in S3 by an isometry, these 120 vertices make up the binary icosahedral group I∗ ⊂ S3. Finally, note that the 600-cell has 600 × 4/2 1200 faces, and 600 × 6/5 720 edges. We summarize our classification of regular polytopes in Table 26.1. To search for new objects, we now go back to Section 16. Recall that we had difficulty realizing certain nonorientable surfaces in R3 because of self-intersections. We will now clarify this. First some examples: Recall that the infinite Mobius¨ band is obtained by continuously sliding and tilting a straight line along a Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

388 26. The Fourth Dimension

Table 26.1 Polytope Schli¨ symbol VE FC

Pentatope {3, 3, 3} 510105

Hypercube {4, 3, 3} 16 32 24 8

16-cell {3, 3, 4} 8243216

24-cell {3, 4, 3} 24 96 96 24

120-cell {5, 3, 3} 600 1200 720 120

600-cell {3, 3, 5} 120 720 1200 600

circle so that in a full round the line completes a half-turn. The motion of the intersection point of the line with the unit circle S1 takes place in the Euclidean plane R2 and is conveniently parameterized by θ → (cos θ, sin θ), θ ∈ R (Figure 26.3). The motion of the line takes place in another copy of R2 perpendicular to this, as shown in Figure 26.4.

Figure 26.3 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

26. The Fourth Dimension 389

θ/2

Figure 26.4

At (cos θ, sin θ) ∈ S1, this line must have slope tan(θ/2) because of the half-turn matching condition. It can be parameterized by

t → (t cos(θ/2), t sin(θ/2)), t ∈ R.

We obtain that the inﬁnite Mobius¨ band can be parameterized by θ θ (θ, t) → cos θ, sin θ, t cos ,tsin ∈ R2 × R2 R4. 2 2 Encouraged by being able to imbed the inﬁnite Mobius¨ band into R4, we now take a look at two other nonorientable surfaces: the Klein bottle K2 and the real projective plane RP2. In Section 16 we realized K2 by rotating a lemniscate in the same way as we rotated a line to obtain the Mobius¨ band. Another (the plumber’s) way of depicting the Klein bottle in R3 is given in Color Plate 11. Apart from the trunk, this picture of K2 is obtained by cutting and pasting pieces of tori and cylinders. These pieces can actually be given by analytical formulas! Slicing K2 along its symmetry plane (with, say, a hacksaw), we obtain two halves of the Klein bottle, one of which is shown in Color Plate 12. Taking a closer look, we see that it is a Mobius¨ band.4 So is the second half, and by pasting we indeed get K2!

4Here is a limerick for the Klein bottle: A mathematician named Klein Thought the M¨obius band was divine. Said he, “If you glue The edges of two You’ll get a weird bottle like mine.” Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

390 26. The Fourth Dimension

Figure 26.5

Remark. ♦ One fine point is worth noting. At the welding circles, the surface is only once differentiable. This is because if a straight line and a circle meet with the same tangent, as in Figure 26.5, then the joined curve is continuously differentiable only up to first order.√ (Indeed, the graph above is given by f :[−1, ∞) → R, f(x) 1 − x2, −1 ≤ x ≤ 0, and f(x) 1, x ≥ 0. Now differentiate at x 0.) This lack of sufficient smoothness is a minor technical problem. As shown in differential topology, there exists an infinitely many times differentiable surface arbitrarily close to the one given.

♣ This representation of K2 in R3 self-intersects in a circle, where the slim 3/4 torus penetrates into the trunk. To get rid of this self- intersection, we now imbed K2(⊂ R3) into R4 using the inclusion R3 ⊂ R4, and then modify the surface locally around this circle so that the modiﬁed surface will not self-intersect in R4. For a good lower dimensional analogy, consider the intersection of the coordinate axes in R2; imbed R2 into R3 and “lift” one axis near the origin. The “lifting” is obtained by introducing the so-called bump-function f : R2 → R, whose graph is depicted in Figure 26.6.

Figure 26.6 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

26. The Fourth Dimension 391

Analytically a bump is given by   1 exp , |p| < 1 f(p) |p|2 − 1  0, |p|≥1.

♦ Notice that f is infinitely many times differentiable even along the unit circle! (Toshow this, differentiate the exponential function at points |p| 1, let |p|→1, and use the fact from calculus that exponential growth is faster than any polynomial growth, that is, x limx→∞ P(x)/e 0 for any polynomial P.) 2 ♣ Bump functions can be “localized.” The function fr : R → R, 2 r>0, defined by fr(p) f(rp)/r, p ∈ R , becomes “supported” on 2 the disk D1/r {p ∈ R ||p| < 1/r} (that is, D1/r is the locus of points on which fr is nonzero). It is now obvious how to steer clear the two axes of R2 in R3. Just take a bump function f and project one axis to the graph of f . This process not only gives us a good clue but tells us what to do to avoid self-intersections of K2 in R4. Just take the model of K2 in R3 with a self-intersection circle assumed to be centered at the origin. Take a bump function f : R3 → R (and notice that we defined f so that it immediately generalizes to any dimensions), localize it on a sufficiently small open ball (so as to avoid interaction with other parts of K2), and project the pipe that penetrates to the trunk 2 4 to the graph of fr. We obtain K in R without self-intersections! This process can be generalized to any compact surface. A fancy way of saying this is that any compact surface can be smoothly imbedded into R4. (Imbedding refers to no self-intersections.)

Remark. Taking this a little further, some apparently strange phenomena emerge when dealing with objects in R4. For example, it is clear what we mean by two circles being “linked” in R3. (Recall from Section 22 that this happens to any two orbits of S1 acting on S3 deﬁning the Hopf map.) Contrary to the evidence demonstrated by some magicians, it is quite obvious that these two circles cannot be unlinked by a continuous motion (homotopy) without breaking one of the circles. The exact proof of this is quite easy once we Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

392 26. The Fourth Dimension

are acquainted with some elements of homotopy theory. On the other hand, if we place the circles (linked in R3)inR4, they can be unlinked by moving one circle away from the coordinate space R3 in the fourth direction. The popularly stated conclusion is that one cannot be chained in R4!

The imbedding result above applies to the real projective plane as well. Instead of repeating the previous argument, we will directly realize RP2—well, not quite in R4, but in S4. Leaving a point out, we can use stereographic projection for S4 to get back to R4. This imbedding will be defined by a map named after Veronese. We first define the Veronese map f : S2 → S4 by 1 1 f(x, y, s) √ (2x2 − y2 − s2), √ (2y2 − x2 − s2), 6 6 1 √ √ √ √ (2s2 − x2 − y2), 3xy, 3xs, 3ys , 6 (x,y,s) ∈ S2 ⊂ R3. This is actually a map from R3 to R6, but we claim that it sends the unit sphere S2 ⊂ R2 given by x2 + y2 + s2 1 to the unit sphere S5 ⊂ R6. Indeed, we work out the sum of squares of the coordinates: |f(x, y, s)|2 (1/6)((2x2 − y2 − s2)2 + (2y2 − x2 − s2)2 + (2s2 − x2 − y2)2) + 3x2y2 + 3x2s2 + 3y2s2 (x2 + y2 + s2)2.

Restricting to spheres, we thus land in S5. Takinga closer look at the ﬁrst three components of f , we see that they add up to zero. Thus the image of f lies in the 5-dimensional linear subspace of R6 given by the normal vector (1, 1, 1, 0, 0, 0) ∈ R6. Think of this linear subspace as a copy of R5. This copy of R5 cuts from S5 ⊂ R6 a copy of the 4-dimensional unit sphere S4! Thus, abusing the notation a bit, we think of the Veronese map being given by f : S2 → S4. The components of f are homogeneous quadratic polynomials in the variables x, y, and s. In particular, f(−x, −y, −s) f(x, y, s) for all (x,y,s) ∈ S2. This means that f takes the same value on Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

26. The Fourth Dimension 393 each antipodal pair of points in S2. It thus defines a map from the quotient RP2 S2/{±I}, so we end up with a map from RP2 to S4. It is now a technical matter to check that this is an imbedding— in particular, it is one-to-one, smooth, and its inverse (from the image to RP2) is also smooth. As noted above, we can now apply stereographic projection to obtain RP2 imbedded into R4. The Roman surface discussed in Section 19 can be obtained from the Veronese map as follows: Project R6 to the linear subspace R3 spanned by the last three coordinate axes. The Veronese map followed by this projection gives a map g : S2 → R3 defined by g(x, y, s) (xy,√ xs, ys), (x,y,s) ∈ S2 ⊂ R3, where we also deleted the coefficient 3 by performing a central dilatation. Since x2 + y2 + s2 1, the components a xy, b xs, c ys satisfy the equation a2b2 + b2c2 + c2a2 abc defining the Roman surface!

Remark. ♦ A more subtle property of the Veronese imbedding is that RP2 sits in S4 as a “soap bubble,” a property that we term minimal. This intuitively means that if we take a small piece of RP2 bounded by a wireframe, the soap ﬁlm in S4 stretched over the frame is given by the Veronese map. Omitting the details, we only hint that minimality is closely connected to the fact that the polynomial components of the Veronese map are harmonic.

♠ Our ambition to understand surfaces such as the Klein bottle and the real projective plane drove us right into four dimensions. The inevitable conclusion is that if we want to do decent mathematics, we have to be able to handle any dimensions. Continuing this multidimensional thought eventually leads us to twenty-first- century mathematics. Since we began these notes with the dawn of mathematical thinking, it is perhaps appropriate to finish it with a note from the nineties. Remember that in Section 17 we classified all finite subgroups of direct spatial isometries. Among these, a prominent role was played by the tetrehedral, octahedral, and Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

394 26. The Fourth Dimension

icosahedral groups. They, or more appropriately, their binary double covers, can be realized as ﬁnite subgroups of quaternions in S3 ⊂ H. Based on analogy with RP2 S2/{±I}, in Section 23 we arrived at the tetrahedral S3/T∗, octahedral S3/O∗, and icosahedral S3/I∗ “manifolds.” Can we view these in some perhaps high-dimensional spheres in the same way as we viewed RP2 in S4? The answer is yes. In fact, DeTurck and Ziller5 showed in 1992 that all these polyhedral manifolds can be imbedded minimally (as spatial soap bubbles) into spheres of large dimension. ♣ We are now getting dangerously close to the research interests of the author6 of these Glimpses. It is time to say goodbye. I hope that you found enough motivation in the sublime beauty of the objects discussed here to carry on with your studies!

Problems

1. Show that the 4-dimensional cube ([0, 1]4 ⊂ R4) can be sliced by 3-dimensional hyperplanes to obtain a 3-dimensional cube, a regular tetrahedron, and a regular octahedron. 2. (a) Show that each edge of the 4-dimensional cube is surrounded by three 3- dimensional cubes. Interpret the Schli¨ symbol {4, 3, 3} of the 4-dimensional cube. (b) Work out the number of vertices, edges, and faces of an n-dimensional cube. 3. (a) Draw several 2-dimensional projections of a 4-dimensional tetrahedron. (b) Show that each edge of the 4-dimensional tetrahedron is surrounded by three 3-dimensional tetrahedra. Interpret the Schli¨ symbol {3, 3, 3} of the 4-dimensional tetrahedron. 4. ♠ Let H denote the linear space of harmonic homogeneous quadratic polynomials in three real variables. Show that the components of the Veronese map span H so that H is 5-dimensional. (With respect to the L2-scalar product on H

5See “Minimal Isometric Immersions of Spherical Space Forms in Spheres,” Comment. Math. Helvetici 67 (1992) 428–458. 6The author determined the modular structure of the “space” of spherical soap bubbles in any dimensions in “Eigenmaps and the Space of Minimal Immersions Between Spheres,” Indiana U. Math. J. 46 (1997) 637– 658. See also the author’s new monograph Finite M¨obius Groups, Minimal Immersions of Spheres, and Moduli, Springer, 2002. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Film 395

defined by integration on S2, the first three components of the Veronese map are the vertices of an equilateral triangle; the last three components form an orthonormal basis in the orthogonal complement of the triangle.) 5. Consider a regular polytope with Schli¨ symbol {a, b, c}. Show that the vertex figures have Schli¨ symbol {b, c}.

Film

T. Banchoff: The Hypercube: Projections and Slicing, Brown University; Interna- tional Film Bureau Inc. (322 S. Michigan Ave., Chicago, IL 60604-4382).

No epilogue, I pray you—for your play needs no excuse. —W. Shakespeare, A Midsummer’s Night’s Dream Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

This page intentionally left blank Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Sets . . . . A . APPENDIX .

In native set theory, the concept of a set is undeﬁned. You may try to say that a set is a family or collection of objects, but these words are just synonyms. We usually denote sets by uppercase Latin letters: A, B, C, etc. If a is an element of A, we write a ∈ A. Otherwise a/∈ A. A set A is contained in a set B,orA is a subset of B, written as A ⊂ B if c ∈ A implies c ∈ B. Two sets A and B are equal, A B,ifA ⊂ B and B ⊂ A. The Cartesian product of two sets A and B is the set A × B consisting of pairs (a, b) with a ∈ A and b ∈ B. The union and intersection of two sets A and B are deﬁned by

A ∪ B {c | c ∈ A or c ∈ B} and

A ∩ B {c | c ∈ A and c ∈ B}.

A map f from a set A to a set B, denoted by f : A → B, is a rule that assigns to each element a ∈ A an element b ∈ B.Ifb is assigned to a, then we write b f(a) and say that b is the value of f on a. The set A is the domain of f , the set B is the range of f , and the image of f is the subset f(A) {b ∈ B | b f(a) for some a ∈ A} of B. 397 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

398 A. Sets

The graph of f is the subset of A × B defined by {(a, f(a)) | a ∈ A}. The map f : A → B is one-to-one if a1 a2 implies f(a1) f(a2). The map f is onto if f(A) B. Finally, f : A → B is a one-to-one correspondence if f is one-to-one and onto. In this case the inverse f −1 : B → A of f exists, with f(a) b if f −1(b) a, a ∈ A, b ∈ B. Two sets have the same cardinality if there is a one-to-one correspondence between them. If A is finite, |A| denotes the number of elements in A. A set is countable if it has the same cardinality as the set of positive integers. The set of all real numbers is not countable. If f : A → B and g : B → C are maps, then the composition g ◦ f : A → C is the map given by (g ◦ f)(a) g(f(a)), a ∈ A. The −1 inverse of f , if it exists, can be defined by saying that f ◦ f IA −1 and f ◦ f IB. Here IA : A → A is the identity map on A—that is, IA(a) a, a ∈ A,—and similarly, IB is the identity map on B. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Groups . . . B . APPENDIX .

An operation on a set A is a map f : A × A → A. It associates to a pair (a1,a2) ∈ A × A the element f(a1,a2) of A. We usually write f(a1,a2) a1 ∗ a2, a1,a2 ∈ A, and call ∗ an operation on A. The operation ∗ on A is associative if

(a1 ∗ a2) ∗ a3 a1 ∗ (a2 ∗ a3), a1,a2,a3 ∈ A, and commutative if

a1 ∗ a2 a2 ∗ a1,a1,a2 ∈ A. A group G is a set on which an associative operation ∗ : G × G → G is given such that 1. There exists a unique element e ∈ G, called the identity, such that e ∗ g g ∗ e g for all g ∈ G; 2. Each element g ∈ G possesses a unique inverse g−1 ∈ G such that g ∗ g−1 g−1 ∗ g e. A group G is called abelian if it is commutative. In speciﬁc groups the operation ∗ is written as addition or multiplication. Additive terminology always assumes that the group is Abelian. If a group G is ﬁnite, then the number of elements |G| in G is called the order of G. 399 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

400 B. Groups

H ⊂ G is a subgroup if the group operation ∗ on G restricts to H, and deﬁnes a group structure on H. The intersection of two subgroups is always a subgroup. Given two groups G and H the Cartesian product G × H carries a group structure. In fact (with obvious notations) we deﬁne (g1 1) ∗ (g2 2) (g1 ∗ g2 1 ∗ h2), g1, g2 ∈ G, h1, h2 ∈ H. A homomorphism f : G → H between groups is a map of the underlying sets such that

f(g1 ∗ g2) f(g1) ∗ f(g2), g1,g2 ∈ G.

The image of a homomorphism f : G → H is always a subgroup of H. The kernel of f : G → H is the subgroup defined by ker (f) {g ∈ G | f(g) e}. A homomorphism is one-to-one iff its kernel is trivial (that is, consists of the identity element alone). A homomorphism that is a one-to-one correspondence between the underlying sets is called an isomorphism. Two groups G and H are called isomorphic, written as G ∼ H, if there is an isomorphism f : G → H. The composition of homomor- phisms is a homomorphism. The inverse of an isomorphism is an isomorphism. Given a group G and a subset G ⊂ G, the smallest subgroup of G that contains G is called the subgroup generated by G; it is denoted by 'G(. In particular, if 'G( G, then G is generated by G;ifG is finite, we say that G is finitely generated. If g ∈ G, the subgroup 'g( consists of all integral powers of g, that is, 'g( {e, g±1,g±2,...} (g0 e). This is called the cyclic subgroup generated by g. If all integral powers of g are distinct, then 'g( is called infinite cyclic.Iftwopowersofg with distinct exponents coincide, then there is a least positive integer n such that gn e. In this case g is said to have order n, and 'g( consists of n elements: 'g( {e,g,...,gn−1}. We say that 'g( is cyclic of order n. Two cyclic groups are isomorphic iff their underlying sets have the same cardinality. Infinite cyclic groups are denoted by the symbol C∞, and finite cyclic groups of order n are denoted by Cn. If G is an abelian group, then the elements of finite order in G form a subgroup called the torsion subgroup Gtor. If, in addition, G Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

B. Groups 401

is ﬁnitely generated, then Gtor is ﬁnite, and we have ∼ r G Gtor × C∞, where r is called the rank of G. A permutation on a set A {a1,...,an} is a one-to-one correspondence f : A → A. A permutation is usually given as

a1 ... an f(a1) ... f(an) We usually assume that A {1, 2,...,n}. The set of all permutations form a group Sn, called the symmetric group on n letters. We have |Sn| n! A subgroup of Sn is called a permutation group. Every finite group is isomorphic to a permutation group. A permutation in Sn is called a transposition if it switches two elements in {1, 2,...,n} and leaves the other elements fixed. Every permutation can be written as the composition of finitely many transpositions. This decomposition is not unique, but the parity (even or odd) of the number of transpositions occuring in the composition is. We say that a permutation is even if it can be written as a composition of an even number of transpositions. Otherwise the permutation is odd. The alternating group An, n ≥ 2, is the subgroup of Sn that consists of all even permutations. We have |An| n!/2. Given a group G and a subgroup H, the left-coset of H by g ∈ G is the subset gH {gh | h ∈ H}. The set of left-cosets gives a partition of the underlying subset of G. The quotient G/H of G by H is the set of all left-cosets: G/H {gH | g ∈ G}.

The number of distinct left-cosets is the index of H in G denoted by [G : H] |G/H|. (This notation usually assumes that G/H is ﬁnite.) G/H is a group under the multiplication (g1H)(g2H) (g1g2)H, g1,g2 ∈ G, iff H is a normal subgroup in G; that is, iff gHg−1 H for all g ∈ G. The kernel of a homomorphism f : G → H is always normal, and the quotient group G/ ker (f) is isomorphic to the image of f . A subgroup of index 2 (such as An in Sn) is always normal, and the quotient is C2. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

402 B. Groups

A group is simple if it has no normal subgroups other than itself and the trivial subgroup (consisting of the identity element alone). The alternating group An is simple for n ≥ 5. A group G is said to act on a set A if, to each element g ∈ G, there is associated a one-to-one correspondence fg : A → A. In addition, −1 f I f −1 (f ) g ∈ G f f ◦ f we require that e A, g g , , and g1g2 g1 g2 , g1,g2 ∈ G. Given an action of G on A, we usually write g instead of fg.IfG acts on A, then the orbit through a ∈ A is the subset G(a) {g(a) | g ∈ G}⊂A.

The isotropy at a ∈ A is the subgroup

Ga {g ∈ G | g(a) a}.

The map that sends the left-coset gGa, g ∈ G,tog(a) establishes a one-to-one correspondence between the quotient G/Ga and the orbit G(a). G acts transitively on G(a) in the sense that any two points in the orbit G(a) can be carried into each other by suitable elements of G. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Topology . . . C . APPENDIX .

Given a set X, a topology on X is a family T {Uα ⊂ X | α ∈ A} of subsets of X indexed by a set A with the following properties:

1. ∪α∈BUα ∈ T for any B ⊂ A; 2. Uα ∩ Uβ ∈ T for any α, β ∈ A; 3. ∅,X ∈ T . The elements of a topology on X are called open subsets of X.If a topology T is given on X, then X is said to be a topological space (with T suppressed). A topological space X is Hausdorff if for any two distinct points x1,x2 ∈ X there exist disjoint open sets U1 and U2 such that x1 ∈ U1 and x2 ∈ U2. (The Hausdorff property is sometimes included in the deﬁnition of topology.) Given a set X,adistance function on X isamapd : X × X → R such that

1. d(x1,x2) ≥ 0, x1,x2 ∈ X, and equality holds iff x1 x2; 2. d(x1,x2) d(x2,x1), x1,x2 ∈ X; 3. d(x1,x2) + d(x2,x3) ≥ d(x1,x3), x1,x2,x3 ∈ X (triangle inequality). A set X with a distance function d is called a metric space. Given a metric space X with a distance function d, the metric ball with 403 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

404 C. Topology

center at x0 ∈ X and radius r>0 is deﬁned as

Br(x0) {x ∈ X | d(x, x0)

A subset U ⊂ X is called open if for every x0 ∈ U there exists ε>0 such that Bε(x0) ⊂ U. In other words, the open sets are unions of metric balls. The triangle inequality ensures that metric balls are open. The open sets form a topology Td on X called the metric topology. The metric topology is always Hausdorff. In the set R of all real numbers, the Euclidean distance between two real numbers x1,x2 ∈ R is d(x1,x2) |x1 − x2|. This defines the Euclidean (metric) topology on R. An open set in R is nothing but the union of (countably many) open intervals in R. Similarly, in the n-dimensional number space Rn the Euclidean distance defines the Euclidean topology in Rn. Given a topological space X with topology T , a subset Y ⊂ X inherits a topology from T by declaring that the open sets in Y are intersections of the elements in T and Y. We say that Y carries the subspace topology. Given two topological spaces X and Y, the Cartesian product X × Y carries a topology by declaring that the open sets in X × Y are unions of sets U × V, where U is open in X and V is open in Y. A subset Y ⊂ X of a topological space X with topology T is called closed if its complement X − Y is open, that is, X − Y ∈ T . Given a subset Y ⊂ X of a topological space X, the closure Y¯ of Y is the intersection of all closed sets in X that contain Y. The closure Y¯ of Y is the smallest closed set that contains Y. In the Euclidean topology, the closure of an open interval (a, b) ⊂ R is the closed interval [a, b] ⊂ R. Amapf : X → Y between topological spaces is continuous at x0 ∈ X if for every open set V in Y that contains f(x0) there exists an open set U in X containing x0 such that f(U) ⊂ V. This generalizes the usual Cauchy definition of continuity of real functions. A function f : X → Y is continuous if it is continuous at every point of X. f : X → Y is continuous iff the inverse image f −1) of any open set V in Y is open in X. A one-to-one correspondence f : X → Y between topological spaces such that f and f −1 are continuous is called a homeomorphism. Two topological spaces are homeomorphic if there is a homeomorphism between them. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

C. Topology 405

Given a topological space X and a partition P of X into mutually disjoint subsets {Xα | α ∈ A}, the family X/P of these disjoint subsets can be made a topological space by requiring that a subset V of X/P be open if the union U of Xα’s that participate in V is open in X. This definition is used to define “pasting” topological spaces as follows: Given two topological spaces X and Y and a homeomorphism f : X1 → Y1 between subspaces X1 ⊂ X and Y1 ⊂ Y, pasting X and Y along f is the quotient X ∪f Y (X ∪ Y)/Pf , where Pf is a partition of X ∪ Y defined as follows: Any point in (X − X1) ∪ (Y − Y1) is a one-element subset of the partition Pf and, for x ∈ X1, {x, f(x)} is a two-element subset in Pf . (In (X ∪ Y)/Pf , the points that are “pasted together” are x and f(x), x ∈ X1.) A topological space X is compact if from any covering of X by open subsets, finitely many members can be extracted that still cover X. A closed subspace of a compact Hausdorff space is compact. The continuous image of a compact topological space is compact. R with respect to the Euclidean topology is noncompact. A subspace X ⊂ Rn is compact iff X is closed and contained in a metric ball. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

This page intentionally left blank Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Smooth Maps . . . D . APPENDIX .

A real-valued function f : U → R of an open set U in Rn is said to be smooth if f possesses continuous partial derivatives up to any order on U.Amapf : U → Rm is smooth if each component of f is smooth. Given any subsets X ⊂ Rn and Y ⊂ Rm,amap f : X → Y is smooth if there exists an open set U ⊂ Rn covering X and a smooth extension of f˜ : U → Rm of f , that is f˜|X f . A smooth map f : X → Y between subsets X ⊂ Rn and Y ⊂ Rm is a diffeomorphism if f is a one-to-one correspondence and f −1 : Y → X is also smooth. X ⊂ Rn and Y ⊂ Rm are diffeomorphic if there is a diffeomorphism f : X → Y. In this case m n. Given a smooth map f : U → V between open sets U ⊂ Rn m and V ⊂ R , the Jacobi matrix of f at x0 is the m × n matrix (f) (∂f i/∂xj)m,n given by Jac x0 i 1,j 1, where the partial derivatives x (f) are evaluated at 0. Multiplication by Jac x0 gives a linear map (df) n → m f x (f) x0 : R R called the differential of at 0.IfJac x0 is nonsingular, then m n and f is a local diffeomorphism at x0. This means that there exists an open set U0 ⊂ U covering x0 and an open set V0 ⊂ V covering f(x0) such that the restriction f |U0 is a diffeomorphism of U0 to V0. The determinant of the Jacobi matrix of f at x0 is called the Jacobian of f at x0. If it is positive, then we say that f is orientation preserving at x0. If it is negative, we say that f is 407 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

408 D. Smooth Maps

orientation reversing at x0.IfJac(f) is nonsingular on a connected open set U, then either f is orientation preserving at every point of U or f is orientation reversing at every point of U. In this case we simply say that f is orientation preserving or orientation reversing. In particular, if f : U → C is a complex function deﬁned on an open set U ⊂ C, and at z0 ∈ U the complex derivative f (z0) (f) f is nonzero, then Jac z0 is nonsingular and is a local diffeomor- 2 phism at z0. Moreover, the Jacobian of f at z0 is |f (z0)| ,sof is orientation preserving at z0. Composition of smooth maps corresponds to composition of their differentials, which in turn corresponds to the product of their Jacobi matrices. By the product theorem for determinants, composition follows the usual parity rule; for example, the composition of two orientation-reversing diffeomorphisms is orientation preserving. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

The . . . . . Hypergeometric . . . . . Differential . . . E . APPENDIX . Equation and the Schwarzian

We summarize1 here some facts on the solutions of second-order homogeneous linear differential equations of the form z p(w)z + q(w), where p and q are rational functions of the complex variable w ∈ C. A point w0 is said to be an ordinary point of the differential equation if p and q have removable singularities at w0, i.e., if w0 is not a pole for p and q. This concept is also extended to w0 ∞as follows. The substitution w˜ 1/w interchanges 0 and ∞, and since dz dz −w˜ 2 , dw dw˜ d2z dz d2z 2w˜ 3 + w˜ 4 , dw2 dw˜ dw˜ 2 the transformed differential equation has the form d2z dz − 2 + 1 p 1 + 1 q 1 z. dw˜ 2 w˜ w˜ 2 w˜ dw˜ w˜ 4 w˜

1For more information, see L. Ahlfors, Complex Analysis, McGraw-Hill, Inc., 1979. 409 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

410 E. The Hypergeometric Differential Equation and the Schwarzian

We say that w0 ∞is an ordinary point of the original equation if w˜ 0 0 is an ordinary point for the transformed equation. This is the case when −(2w + w2p(w)) and w4q(w) have removable singularities at w0 ∞. If w0 is an ordinary point of our differential equation, then p and q are analytic at w0, and hence there exists a local solution z (de- ﬁned on an open neighborhood of w0) with arbitrarily prescribed z(w ) z z(w ) z z values 0 0 and 0 0. The local solution is analytic at w0, and its germ at w0 is uniquely determined. (Indeed, a simple induction in the use of the Cauchy inequality shows that a formal z ∞ a (w − w )n power series expansion n 0 n 0 has positive radius of convergence.) Analytic continuation along curves that avoid the poles of p and q extends any local solution to a global solution. The global solution is, in general, multiple-valued. We now begin to study the behavior of the solutions of our differential equation near a singular point w0. We ﬁrst assume that w0 is the simplest possible singularity in the sense that both p and q have only simple poles at w0. For simplicity, we assume that w0 0. (This can always be attained by replacing w − w0 with w.) We write the Laurent expansions of p and q at w0 0 as follows: p p(w) −1 + p + p w +···, w 0 1 q q(w) −1 + q + q w +···. w 0 1

z ∞ a wn Once again, if we substitute the formal power series n 0 n into the differential equation, we ﬁnd that only a0 can be chosen arbitrarily, and the recurrence formula for the coefﬁcients an works only under the condition that p−1 ∈ {0, 1, 2,...}. With this restriction there is always an analytic solution in a neighborhood of w0. The most general case for which this process gives an explicit solution is that in which w0 is a regular singular point of the differential equation, i.e., when p has at most a simple pole at w0, and q has at most a double pole at w0. Assume that w0 is a regular α singular point. We seek a solution of the form z (w − w0) g(w), where g is analytic and nonzero at w0. Once again, setting w0 0 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

E. The Hypergeometric Differential Equation and the Schwarzian 411 and substituting, we ﬁnd that g satisﬁes the differential equation α αp(w) α(α − ) g p(w) − 2 g + q(w) + − 1 g. w w w2

If we choose α to satisfy the indicial equation

α(α − 1) p−1α + q−2, where q q q(w) −2 + −1 + q + q w +···, w2 w 0 1 then in the differential equation for g, the coefﬁcients have at most simple poles at w0 0, so that the previous discussion applies. We obtain that the original differential equation has a solution of α the form z w g(w), where g is analytic and nonzero at w0 0, provided that p−1 − 2α ∈ {0, 1,...}. Here α is one of the two indicial roots α1,α2, solutions of the indicial equation. Since α1 + α2 α p−1 + 1, the condition for the existence of a solution z w g(z) as above for α α1 is equivalent to α2 − α1 ∈ {1, 2,...}. Switching the roles of α1 and α2, we see that our differential equation has α1 two linearly independent solutions of the form z1 w g1(w) and α2 z2 w g2(w), provided that α1 − α2 is not an integer.

Theorem. If w0 is a regular singular point of the differential equation z p(w)w + q(w), then there exist linearly independent solutions of the form

α1 α2 (w − w0) g1(w) and (w − w0) g2(w), where α1,α2 are the solutions of the indicial equation

α(α − 1) p−1α + q−2, and g1,g2 are analytic and nonzero at w0, provided that α1 − α2 is not an integer. If α2 − α1 is a nonnegative integer, then there is always one α1 solution of the form (w − w0) g1(w), where g1 is analytic and nonzero at w0. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

412 E. The Hypergeometric Differential Equation and the Schwarzian

A simple analysis shows that our differential equation with one or two regular singular points (and all other points ordinary) can be explicitly solved. Toget something new, we thus assume that we have three regular singular points. Since a linear fractional transformation does not change the structure of the differential equation (including the character of the singularities), we may assume that the three regular singular points are 0, 1, and ∞ (in fact, by the results of Section 12, any three points in the extended complex plane Cˆ can be mapped to 0, 1, and ∞ by a suitable linear fractional transformation). Since 0 and 1 are regular singular points (and all other points are ordinary), we must have A B p(w) + + P(w), w w − 1 C D E F q(w) + + + + Q(w), w2 w (w − 1)2 w − 1 where P and Q are polynomials. Since the singularity at ∞ is also regular, 2w + w2p(w) must have at most a simple pole at ∞, and w4q(w) must have at most a double pole at ∞. The former condition amounts to the vanishing of P, the latter to the vanishing of Q and D + F. We obtain A B p(w) + , w w − 1 C D E q(w) − + . w2 w(w − 1) (w − 1)2

We now bring in the indicial equations. At w0 0 the indicial equation is α(α − 1) Aα + C.

Letting α1,α2 denote the indicial roots, we have A α1 + α2 − 1 and C −α1α2. Similarly, at w0 1, the indicial equation is β(β − 1) Bβ + E.

If β1,β2 are the roots, then B β1 + β2 − 1 and E −β1β2. 2 Finally, since at w0 ∞the leading coefﬁcients of −2w − w p(w) and w4q(w) are −(2 + A + B) and C − D + E, the indicial equation Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

E. The Hypergeometric Differential Equation and the Schwarzian 413 at w0 ∞is

γ(γ − 1) −(2 + A + B)γ + C − D + E.

If γ1,γ2 are the roots, then we have γ1 + γ2 −A − B − 1 and γ1γ2 −C + D − E. Expressing A,B,C,D,Ein terms of the six roots, our differential equation takes the form − α − α − β − β z + 1 1 2 + 1 1 2 z w w − 1 α α α α + β β − γ γ β β + 1 2 − 1 2 1 2 1 2 + 1 2 z 0, w2 w(w − 1) (w − 1)2 and

α1 + α2 + β1 + β2 + γ1 + γ2 1.

This is called the hypergeometric differential equation. The solutions of the hypergeometric differential equation are called hypergeometric functions. By construction, the hypergeometric differential equation is the canonical form of a homogeneous second-order differential equation with exactly three regular singular points (and ordinary points elsewhere). The hypergeometric differential equation is often written in the equivalent form

z z − (1 − α − α ) − (1 + γ + γ )w w(w − 1) 1 2 1 2 z + α α − (α α − β β + γ γ )w + γ γ w2 0. w2(w − 1)2 1 2 1 2 1 2 1 2 1 2

From now on, we assume that none of the differences α1 − α2, β1 − β2, γ1 − γ2 is an integer. To simplify and at the same time solve the hypergeometric differential equation, we recall that the substitution wαg(w) determines for g a similar differential equation. As a simple computation shows, the indicial roots of the differential equation for g are α1 − α and α2 − α, and the indicial roots at ∞ are γ1 + α and γ2 + α. Introducing (somewhat prematurely) Riemann’s Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

414 E. The Hypergeometric Differential Equation and the Schwarzian

notation, we express this as      01∞   01∞  P α β γ ,w zαP α − αβ γ + α ,w.  1 1 1   1 1 1  α2 β2 γ2 α2 − αβ2 γ2 + α

Here the Riemann P-function on the left-hand side stands for all locally deﬁned solutions of our differential equation, and the equality means that any local solution is equal to a solution of the differential equation for g multiplied by zα, and the indicial roots undergo the changes as indicated. (We are not going into the proper interpretation of relations like this; in particular, we will not discuss unicity in any detail. We ﬁnd, however, this notation undeniably suitable for tabulating the indicial roots.) We can also separate the factor (w − 1)β from g and obtain    01∞  P α β γ ,w  1 1 1  α2 β2 γ2    01 ∞  wα(w − 1)βP α − αβ− βγ+ α + β ,w.  1 1 1  α2 − αβ2 − βγ2 + α + β

Up to this point, α and β did not have any preassigned values. We α1 β1 now set α α1 and β β1. After separating the factor w (w−1) , the remaining functions    01 ∞  P 00γ + α + β ,w  1 1 1  α2 − α1 β2 − β1 γ2 + α1 + β1

satisfy the hypergeometric differential equation c 1 − c + a + b ab z + + z + z 0. w w − 1 w(w − 1) Here we used classical notation

a α1 + β1 + γ1,b α1 + β1 + γ2,c 1 + α1 − α2, Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

E. The Hypergeometric Differential Equation and the Schwarzian 415 noting also that

c − a − b β2 − β1.

Multiplying out, we can write this differential equation as

w(1 − w)z + (c − (a + b + 1)w)z − abz 0.

z ∞ A wn This equation has a solution in the form n 0 n . In fact, substituting we ﬁnd an embarrassingly simple recurrence relation that, up to a constant multiple, gives the solution a · b a(a + 1) · b(b + 1) F(a, b, c, w) 1 + w + w2 1 · c 1 · 2 · c(c + 1) a(a + 1)(a + 2) · b(b + 1)(b + 2) + w3 +···. 1 · 2 · 3 · c(c + 1)(c + 2)

In a similar vein, setting α α2 and β β1, after separating the α β factor w 2 (w − 1) 1 , the remaining functions are    01 ∞  P α − α 0 γ + α + β ,w.  1 2 1 2 1  0 β2 − β1 γ2 + α2 + β1

Rewriting this in terms of the parameters a, b, c, we ﬁnd that w1−cF(1 + a − c, 1 + b − c, 2 − c, w) is another solution linearly −c α −α independent of F(a, b, c, w). (The factor w1 w 2 1 makes up α β α β the quotient w 2 (w − 1) 1 /w 1 (w − 1) 1 .) Linearly independent solutions near 1 can be found using the identity      01∞   01∞  P α β γ ,w P β α γ , 1 − w .  1 1 1   1 1 1  α2 β2 γ2 β2 α2 γ2

Solutions near ∞ can be written down in an analogous way. Summarizing, we ﬁnd that up to various multiplicative powers of the independent variable (depending on the indicial roots at 0, 1, and ∞), all solutions of the hypergeometric differential equation can be written in terms of the hypergeometric functions F. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

416 E. The Hypergeometric Differential Equation and the Schwarzian

We now return to the general setting and consider again our differential equation z p(w)z + q(w)z, where p and q are rational functions. Given two linearly independent but possibly multiple-valued solutions z1 and z2, any solution of the differential equation can be written as a linear combination of z1 and z2. Consider the ratio z z1/z2. (We are overusing the symbol z a little here. Whether z stands for the ratio z1/z2 or for the general variable in our differential equation will be clear in the text.) If instead of z1 and z2 we choose a different pair of linearly independent solutions, then the corresponding ratio will be a linear fractional transformation applied to z. Indeed, if z˜1 az1 + bz2 and z˜2 cz1 + dz2 (a, b, c, d ∈ C, ad − bc 1), then z˜ ˜z1/z˜2 (az + b)/(cz + d).

Remark. Considering closed paths based at a fixed point (and away from the poles of p and q), analytic continuation of solutions along these paths give rise to the concept of monodromy, a homomorphism of the fundamental group of Cˆ minus the poles of p and q into SL(2, C). In 1873 Schwarz classified all hypergeometric differential equations with finite monodromy.

This motivates us to seek a differential expression S that remains invariant under all linear fractional transformations. Among the four quantities a, b, c, d there is a relation ad − bc 1, so that S must be a third-order differential operator. An explicit expression for S can be derived in complete generality as follows. We let z be an analytic function and z˜ (az + b)/(cz + d). Suppressing the variable w ∈ C, we write this as czz˜ + dz˜ − az − b 0. Differentiating, we obtain c(zz˜ + zz˜) + dz˜ − az 0, c(zz˜ + 2zz˜ + zz˜) + dz˜ − az 0, c(zz˜ + 3zz˜ + 3zz˜ + zz˜) + dz˜ − az 0. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

E. The Hypergeometric Differential Equation and the Schwarzian 417

We view these three equations as a linear system for a, c, and d. Its determinant, after canceling the highest-order derivative mixed terms by column operations, is

z˜ z 0 − zz˜ z˜ z . 2 3zz˜ + 3zz˜ z˜ z

This determinant must vanish, since ad − bc 1. Expanding and grouping similar terms together, we obtain z˜ z˜ 2 z z 2 − 1 − 1 . z˜ 2 z˜ z 2 z This tells us that S should be defined as z z 2 S(z) − 1 . z 2 z We call S(z) the Schwarzian of z. By our computations, the Schwarzian is invariant under linear fractional transformations. Conversely, if the Schwarzians of two analytic functions coincide, then the two functions differ by a linear fractional transformation. This follows from unicity of solutions of third-order differential equations. We now return to the main line. If z1 and z2 are linearly independent solutions of our differential equation, then the Schwarzian S(z) of the ratio z z1/z2 will not depend on the particular choice of z1 and z2, but only on the coefficients p and q. To work out this dependence explicitly, we fix a choice for z1 and z2. Since they both satisfy our differential equation, we have zz − z z p(w)(z z − z z ). 1 2 1 2 1 2 1 2 Using this, we work out the ingredients of the Schwarzian derivative of z. We have z z − z z z 1 2 1 2 . z2 2 Logarithmic differentiation gives z zz − z z z (ln z) (ln(z z − z z )) − 2(ln z ) 1 2 1 2 − 2 2 . z 1 2 1 2 2 z z − z z z 1 2 1 2 2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

418 E. The Hypergeometric Differential Equation and the Schwarzian

Combining this with the above, we obtain z z p − 2 . 2 z z2 Finally, we compute z 1 z 2 z 1 z 2 S(z) − − −p + 2 − −p + 2 2 2 z 2 z z2 2 z2 1 z − pz 1 p − p2 − 2 2 2 p − p2 − 2q. 2 z2 2

Summarizing, we obtain that if z1 and z2 are linearly independent solutions of our differential equation, then z z1/z2 has Schwarzian derivative 1 S(z) p − p2 − 2q. 2 We see that to every third-order differential equation of the form S(z) s,

with s a rational function, there corresponds a homogeneous second-order linear differential equation via 1 s p − p2 − 2q. 2 The rational function s obviously does not determine the coefﬁ- cients p and q uniquely. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Galois Theory . . . F . APPENDIX .

We review here some basic ingredients of Galois theory.1 A field K is a set equipped with two operations: + (addition) and × (multiplication), such that K is an abelian group with respect to the addition, K −{0} is an abelian group with respect to the multiplication, and the distributive law holds in K. The prime subfield of a field K is the intersection of all subfields of K. The prime subfield is either isomorphic to the field Q of rational numbers, or the field Zp,for some prime p. In the former case, we say that K has zero characteristic, in the latter that K has characteristic p. All fields considered here will be of characteristic zero. Given a field extension k ⊂ K (k is a subfield of K), written as K/k, the degree [K : k]ofK/k is the dimension of K as a linear space over k.Ifk ⊂ L ⊂ K is a chain of field extensions, then [K : k] [K : L][L : k]. We say that the field extension K/k is finite if [K : k]is finite. For example, a field extension K/k is quadratic if [K : k] 2. A quadratic√ field extension K/k can√ be obtained by adjoining a square root√ δ to k, where δ ∈ k but δ ∈ k; we write the extension as K k( δ).Ak-automorphism of K is an automorphism of K that

1For more details, see M. Artin, Algebra, Prentice Hall, 1991, or I. Stewart, Galois Theory, Chapman and Hall, 1973. 419 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

420 F. Galois Theory

fixes k. The Galois group G G(K/k) of the extension K/k is the group of k-automorphisms of K.IfK/k is a finite field extension, then the Galois group G is finite. For any finite field extension K/k, the order |G| of G divides [K : k]. We call K/k a Galois extension if |G| [K : k]. If K/k is Galois then the fixed field KG is equal to k. Conversely, if G is a finite group of automorphisms of a field K and k KG denotes its fixed field, then K/k is a Galois extension with Galois group G. Given a Galois extension K/k with Galois group G, there is a one-to-one correspondence between the subgroups H of G and the intermediate fields k ⊂ L ⊂ K. The correspondence is given by L KH and H G(K/L). We also have [K : L] |H| and [L : k] [G : H]. In addition, L/k is Galois iff H is normal in G, and in this case, G(L/k) G/H. Let k[z] denote the ring of polynomials with coefficients in k in the variable z. Given a nonconstant monic polynomial P ∈ k[z], a splitting field for P over k is an extension field K such that P factors into linear factors in K, and K is the smallest extension of k with this property. A splitting field K is generated over k by the roots of P.A splitting field always exists and is unique up to a k-isomorphism, an isomorphism between the splitting fields that fixes the common ground field k. A splitting field over k is always a Galois extension of k. Conversely, any Galois extension of k is the splitting field of some polynomial P ∈ k[z]. If P ∈ k[z] is irreducible, then the Galois group G of the splitting field of P is uniquely determined by P.For this reason, we say that G is the Galois group of P. Given a splitting field K of a polynomial P ∈ k[z] of degree n over k with roots z1,...,zn, the Galois group G acts on the roots faithfully, so that G can be thought of as a subgroup of the symmetric group Sn on n letters. The action on the roots is transitive iff P is irreducible over k. The discriminant δ of P is 2 δ (zj − zl) . 1≤j

F. Galois Theory 421 √ √ G(K/k( δ)) ⊂ An, so that by adjoining δ to k (at the expense of a quadratic extension) we can make G a subgroup of An. Let K/k be a Galois extension with Galois group G. Given z∗ ∈ K − k, consider the orbit G(z∗) {z∗,...,z∗ } 1 n∗ and the polynomial n∗ ∗ ∗ P (z) (z − zj ). j 1 Then P∗ ∈ k[z] is irreducible and [k(z∗) : k] n∗. In particular, n∗ divides |G|.IfK is the splitting ﬁeld of a polynomial P ∈ k[z] with ∗ roots z1,...,zn, then z is a rational function of the roots z1,...,zn. In this case we call P∗ a resolvent polynomial of P.

Example Let K be the splitting ﬁeld of an irreducible quartic P ∈ k[z] with roots z1,z2,z3,z4. Let z∗ (z + z )(z + z ), 1 1 2 3 4 z∗ (z + z )(z + z ), 2 1 3 2 4 z∗ (z + z )(z + z ). 3 1 4 2 3 These elements are distinct, since P is irreducible (cf. the argument at the end of Section 6). The symmetric group S4 acts on {z∗,z∗,z∗} G ⊂ S 1 2 3 transitively. If the Galois group 4 is also transitive {z∗,z∗,z∗} on 1 2 3 , then the resolvent cubic 3 ∗ ∗ P (z) (z − zj ) j 1 √ P k G A S δ ∈ k of√ is irreducible over , and is 4 or 4 according as or δ/∈ k. Let w ∈ C, and consider the splitting ﬁeld of the polynomial P(z) zn − w, where n ≥ 2 is an integer. We choose k to be generated over Q by w and the primitive nth root of unity ω e2πi/n. The complex roots Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

422 F. Galois Theory

of P are the nth roots of w, and if z∗ denotes one of them, then the roots of P can be listed as z∗,z∗ω, z∗ω2,...,z∗ωn−1.

It follows that the splitting ﬁeld K of P is generated over k by a single root: K k(z∗).Ifw is not an nth power in k, then [K : k] n, and the Galois group G(K/k) is a cyclic group of order n.

Theorem. Let n ≥ 2 be an integer. Let k be a ﬁeld and assume that k contains the primitive nth root of unity ω e2πi/n. Assume that K is a Galois extension of k with Galois group G, a cyclic group of order n. Then there exists w∗ ∈ k such that K is the splitting ﬁeld of the polynomial zn − w∗, and K is generated by any of the roots. Moreover, for a given root z∗ ∈ K, K k(z∗), there is an isomorphism

ϕ : G → Cn such that σ−1(z∗) ϕ(σ)(z∗), σ ∈ G.

Remark. ∗ n ∗ ∗ We call a solution z ∈ K of√ the equation z w , w ∈ k,aradical over k. We also write z∗ n w∗.

In general, let P ∈ k[z] and let K be the splitting ﬁeld of P with Galois group G. Consider a composition series of G:

G G0 ⊃ G1 ⊃ ··· ⊃ GN {1}.

Here, each Gm+1 is a maximal normal subgroup in Gm, and thus, the factor Gm/Gm+1 is simple, m 0,...,N − 1. The composition series of G is not unique, but by the Jordan–Holder¨ theorem, up to order, the factor groups are. Consider the corresponding chain of ﬁeld extensions

k k0 ⊂ k1 ⊂ ··· ⊂ kN K,

where km is the fixed field of Gm, m 1,...,N − 1. Since Gm+1 is normal in Gm, km+1 is a Galois extension of km with Galois group Gm/Gm+1. Being a Galois extension, km+1 can be thought of as the splitting field of a polynomial Pm over km. We see that finding the Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

F. Galois Theory 423 roots of P amounts to solving the polynomial equations Pm(z) 0, for all m 0,...,N − 1. Assume now that G is solvable, that is, each factor Gm/Gm+1 p |G /G | is abelian, thereby cyclic of (prime) order m m m+1√.By pm the theorem above, km+1 is generated over km by a radical wm, wm ∈ km. Applying this procedure inductively, we obtain that every element of K (including the roots z1,...,zn of P) can be expressed by radicals over k (in the obvious sense). Since the converse is clear, we see that the roots of a polynomial P ∈ k[z] are expressible by radicals over k iff the Galois group of the splitting ﬁeld of P over k is solvable. All subgroups of S4 are solvable. It follows that the roots of polynomials of degree 4 or less are expressible by radicals. We derived the explicit root formulas in Section 6. In contrast, an irreducible quintic with Galois group A5 has no radical formula, since A5 is simple. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

This page intentionally left blank Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Solutions for 100 . . . . . Selected Problems . . .

Section 1.

1. (a) Write the number in the form 3k,3k + 1, or 3k + 2, take the squares of these integers, and consider the remainders modulo 3. (b) Write the number in the form 7k,7k ± 1,7k ± 2, or 7k ± 3, take the cubes of these, and consider the remainders modulo 7. 2. By Problem 1(b), we need to show that 7a − 1 is not a square. This follows as in 1(b) by squaring the numbers 7k,7k ±1,7k ±2, and 7k ± 3, and studying the remainders modulo 7. n−1 n−2 4. Let a a110 + a210 + ··· + an, where a1,a2,...,an ∈ {0, 1,...,9}, a1 0. Write 10 9 + 1, and use the binomial theorem to expand the powers of 10 in the expression for a. 5. Since Zp is a ﬁeld, the statement is equivalent to the vanishing of

p−1 p−1 p−1 [1/a2] [a]−2 [a¯]2, a 1 a 1 a 1

where [a¯] is the inverse of [a]inZp. Now notice that the last sum is the sum of squares of the nonzero elements in Zp. 425 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

426 Solutions for 100 Selected Problems

Section 2. √ n n 1. a is a rational solution of the equation x a. ∞ (− )k/( k + ) 2. Write cos 1 k 0 1 2 1 ! and repeat, with appropriate modiﬁcations, the proof of irrationality of e as in the remark following Theorem 1. 3. If a/b is a root of P(x) 8x3 − 6x − 1, then a ±1 and b|8. Now check all possible combinations. 4. Using the geometric series formula, a Riemann sum can be worked out as m m α − 1 (αk)n(αk − αk−1) α(n+1)k α k 1 k 1 αm(n+1) − 1 (α − 1)αn α(n+1) − 1 an+1 − 1 . αn + αn−1 +···+α + 1 √ For m →∞, we have α m a → 1, so that the Riemann sum converges to an+1 − 1 . n + 1 5. This is a primary example for induction. For another proof, differentiate the geometric series formula with respect to x, d 1−xn+1 x + +···+n x→ substitute 1 and obtain 1 2 lim 1 dx 1−x . To evaluate the limit, use L’Hospital’s rule. 7. As in Problem 5, differentiate the geometric series formula with respect to x twice and substitute x 1. 8. The pyramid staircase can be cut into a large square pyramid of volume n3/3, and two sets each consisting of n triangular prisms of heights 1, 2,...,n, common base area 1/2, and total volume (1 + 2 +···+n)/2. Notice that the two sets of prisms overlap in n small square pyramids with total volume n/3, and this has to be subtracted from the volume. The volume of the pyramidal staircase is thus n3/3 + (1 + 2 + ...+ n) − n/3. Now use Problem 5. a N c n c ∈{ , } c a 10. Write n 0 n2 with n 0 1 . Notice that n 1 iff n is odd. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Solutions for 100 Selected Problems 427

11. (c) 1/ Tri (n) 2(1/n − 1/(n + 1)).

Section 3. 2. Split the triangle into 3 subtriangles by drawing line segments from the center of the inscribed circle to the vertices. The sum of the areas of the subtriangles, ar/2 + br/2 + cr/2, must be equal to the area of the original triangle, ab/2. We obtain r ab/(a + b + c). Without loss of generality we may assume that a, b, c are relatively prime. Since (a,b,c) is Pythagorean, we have a 2st, b t2 − s2, and c t2 + s2. Substitute these into the expression of r and simplify. 3. We may assume that a, b, c are relatively prime. Let a 2st, b t2 − s2, and c t2 + s2 with t>srelatively prime and of different parity. Since 2|a and 2|st,wehave4|a. Write ab 2st(t−s)(t+s) and notice that 3 must divide one of the numbers s, t, t − s,ort + s. (This follows by considering the remainders of t and s modulo 3). Thus 12|ab. Similarly, 5 divides one of the numbers s, t, t − s, t + s,ort2 + s2. (This follows again by looking at the remainders of s and t modulo 5.) 4. For a, b, c consecutive, we have a + c 2b. Substituting this into the Pythagorean equation, we obtain 4a 3b. Since a, b are relatively prime, a 3 and b 4 follow. 5. We claim that there is no solution a for b ≥ 2. Notice ﬁrst that a and a + 2 have the same parity, so that a + 1 must be even, say, a + 1 2k. Substituting, we have (2k − 1)2b + (2k)2b (2k + 1)2b. Expanding both sides by the binomial formula and grouping, conclude that k must divide b. On the other hand, dividing all terms by (2k)2b, we have

b b 1 2 1 2 1 − + 1 1 + ; 2k 2k in particular, the estimate b 1 2 1 + < 2 2k Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

428 Solutions for 100 Selected Problems

holds. Use the binomial formula again to conclude that b/k < 1. This contradicts k|b. 6. If (a,b,c)is a Pythagorean triple with a + b + c ab/2, then (a − 4)(b − 4) 8. 7. For n ≥ 4, let a n(n2 − 3) and b 3n2 − 1. 8. (a) (0, 0) is a double point; y2 x2(x + 2). Setting y mx, m ∈ Q , we obtain x m2 − 2 and y m(m2 − 2). (b) (−1, 0) is a double point; y2 (x + 1)2(x − 2). Setting y m(x + 1), m ∈ Q , we obtain x m2 + 2 and y m(m2 + 2). 9. Let (x0,y0), y0 0, be a rational point on the Bachet curve given by y2 x3 + c. Implicit differentiation gives 2yy 3x2, m (x ,y ) m x2/ y so that the slope of the tangent line at 0 0 is 3 0 2 0. We need to couple the equation of the tangent line y − y0 2 3 m(x − x0) with Bachet’s equation y x + c to obtain the y2 x3 + c coordinates of the intersection point. Since 0 0 ,we y2 − y2 x3 − x3 write Bachet’s equation as 0 0. After factoring, (y − y )(y + y ) (x − x )(x2 + xx + x2) we have 0 0 0 0 0 . Substituting y − y0 m(x − x0) and canceling the factor x − x0, we obtain m(y + y ) x2 + xx + x2. 0 0 0

Substituting again y − y0 m(x − x0), we ﬁnally arrive at m( y + m(x − x )) x2 + xx + x2. 2 0 0 0 0

This is a quadratic equation in x that has x0 as a root, since m x2/ y 3 0 2 0. Factoring, we obtain x4 − 8cx (x − x ) x − 0 0 0. 0 y2 4 0 Bachet’s formula follows. 10. (1, 1) is a rational point. Write the equation of the circle as (x2 − 1) + (y2 − 1) 0 and use y − 1 m(x − 1), m ∈ Q , when factoring. 12. (a) a2 2b2 implies that (2b − a)2 2(a − b)2. 14. 78 2 · 3 · 13 ab/2 and a2 + b2 c2. In particular, a, b are relatively prime, since 62 + 262 712 is not a square. Setting a 2st, b t2 − s2, and c t2 + s2, we obtain 2 · 3 · 13 st(t − s)(t + s) with t>srelatively prime and of different parity. This is impossible. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Solutions for 100 Selected Problems 429

Section 4.

1. Multiply the polynomial by the least common multiple of the denominators of the coefficients. 3.√ If c is a root of a polynomial P with integer coefficients, then c is a root of the polynomial Q , where Q(x) P(x2). 4. Let P and Q be polynomials of degrees m and n with rational coefficients and leading coefficient one such that P(c1) Q(c2) P Q c α ,...,α c β ,...,β 0. Let and have roots 1 1/ /m and 2 1 n. R(x) m n (x − (α + β )) Consider the polynomial i 1 j 1 i j . The coefficients of R are symmetric in αi and βj, and therefore, by the fundamental theorem of symmetric polynomials, they can be expressed as polynomials (with integral coefficients) in the coefficients of P and Q . In particular, the coefficients of R are R(c + c ) c + c rational. Since 1 2 0, 1 2√is algebraic. √ 5. An isomorphism would associate√ to 3 an element a + b 2 and thus to 3 the√ element (a + b 2)2. Since√ 3 1 + 1 + 1in ( ) (a + b )2 both fields, in Q √2 we have 3 √ 2 . 7. Since (x − (a + b p))(x − (a − b p)) x2 − 2ax + a2 + b2p, the remainder is a polynomial of degree less than or equal to 1 with rational coefficients. 10. The existence and uniqueness of x0 follow from the properties of the exponential function. Let m be rational. If x0 is rational, x0 then so is e mx0 + 1, and this contradicts the corollary to Theorem 1 in Section 2. Transcendentality of x0 for algebraic m follows from Lindemann’s theorem.

Section 5.

1. The equation of the circle is a − c 2 b 2 (a − c)2 + b2 x − + y + . 2a 2a 4a2 Setting x 1 and solving for y, we obtain the quadratic formula for r1 and r2. A generalization of this construction to cubic polynomials would mean that the roots of a cubic polynomial with constructible coefﬁcients would be constructible. This is false (cf. Problem 4 in Section 6). Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

430 Solutions for 100 Selected Problems √ 3 −i 3 z( π/ ) z(π/ + kπ/ ) k , , i 2. √ 3 2 2 2 3 , 0 1 2. These give and ± 3/2 − i/2. 3. The two lines are parallel iff (z1 − z2)/(z3 − z4) is real. Thus, we may assume that the two lines intersect. The condition ! (z1 − z2)/(z3 − z4) 0 is invariant under translation (which amounts to adding a constant to each variable), so that we may assume that the lines intersect at the origin. Then z1,z2 and z3,z4 are real constant multiples of each other, and the condition reduces to !(z1/z3) 0. Now write z1 and z3 in polar form and consider the argument of the ratio z1/z3. 5. The condition remains invariant when we translate, rotate, and scale each triangle individually. (For example, translation corresponds to the row operation of adding a constant multiple of the ﬁrst row to the row representing the triangle.) We can thus reduce the problem to the case where z1 w1 0 and z w z w 2 2 1.√ The condition√ now reduces to 3 3. 6. 2 · 3 (1 + 5i)(1 − 5i). 7. Letting z1 a1 +b1i, and z2 a2 +b2i, a1,a2,b1,b2 ∈ N, we have |z |2|z |2 (a2 +b2)(a2 +b2) |z z |2 (a a −b b )2 +(a b + 1 2 1 1 2 2 1 2 1 2 1 2 1 2 2 a2b1) . Any integer can be written as a power of 2 multiplied by numbers of the form 4k + 1 and 4k + 3. Now apply Fermat’s theorem and the above.

Section 7. 1. (a) N (0, 0, 1), p (a,b,c), and hN (p) a/(1 − c), b/(1 − c), −1 0 are collinear. (c) The stated expression for hN (z) has unit −1 length, and N, z, and hN (z) are collinear. 2. As in Problem 1, for p (a,b,c) ∈ S2, we have

a + bi hS(p) ∈ C. 1 + c

Moreover, z |z|2 − −1 2 1 2 hS (z) , − ∈ S ,z∈ C. |z|2 + 1 |z|2 + 1 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Solutions for 100 Selected Problems 431

Hence, z/(|z|2 + ) z −1 2 1 N ◦ hS )(z) . 1 + (|z|2 − 1)/(|z|2 + 1) |z|2 4. We have |z − w|2 (z − w)(z¯ − w)¯ |z|2 − zw¯ −¯zw +|w|2

and |1 − zw¯ |2 (1 − zw)(¯ 1 −¯zw) 1 − zw¯ −¯zw +|z|2|w|2. The stated inequality is thus equivalent to 0 <(1 −|z|2)(1 − |w|2).

Section 8. 4. (a) If P is an odd-degree polynomial with real coefficients, then limx→±∞ P(x) ±sgn (cn) ∞, where sgn (cn) is the sign of the leading coefficient cn of P. By continuity, P must have a real zero. (b) The polynomial Qk, k ∈ Z, has coefficients that are, up to sign, the elementary symmetric polynomials in the variables αi + αj + kαiαj,1≤ i

432 Solutions for 100 Selected Problems

quadratic polynomial (x − z)(x −¯z) x2 − 2!(z)x +|z|2 and study the remainder.

Section 9. 1. The regular hexagon is obtained by slicing the cube with a plane through the origin with normal vector (1, 1, 1). The plane extensions of the three sides of the cube meeting at (1, 1, 1) cut an equilateral triangle out of this plane, and the plane extensions of the other three sides of the cube meeting at (−1, −1, −1) further truncate this triangle to a regular hexagon. 2. Tv ◦ Rθ(p) ◦ T−v is a direct isometry, and it leaves q ﬁxed. Thus it must be a rotation with center at q. The translations do not change the angle. 3. (a) The vertices of P2n are those of Pn plus the midpoints of the circular arcs over the sides of Pn. Thus, half of a side of Pn and P a side of 2n meeting at a vertex of Pn are two sides of a right triangle whose third side is 1 − 1 − (sn/2)2. We thus have   2 2 2 sn sn s2 + 1 − 1 −  . 2n 2 2

This gives 2 s2n 2 − 4 − sn. √ In particular, since s4 2, we have √ s2n 2 − 2 + 2 +···+ 2,

with n − 1 nested square roots. (b) Let An denote the area of Pn. Since half of the sides of Pn serve as heights of the 2n isosceles triangles that make up P2n, we have A2n nsn/2. In particular, we have √ n−1 n−1 A2n+1 2 s2n 2 2 − 2 + 2 +···+ 2,

with n − 1 nested square roots. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Solutions for 100 Selected Problems 433

5. (a, b) is the normal vector of the reciprocal line. Thus, by symmetry, it is enough to consider the reciprocal of z(0) 1. The radius of the circle inscribed in Pn is r |1 + z(2π/n)|/2, and the line reciprocal to 1 is x r2. Now the statement follows from the identity !(1 + z(θ))/2 |1 + z(θ)|2/4.

Section 10. 2. Type 5. 3. Let L {kv + lw | k, l ∈ Z} be a lattice. A half-turn about the midpoint of the lattice points k1v + l1w and k2v + l2w sends a point p to (k1 + k2)v + (l1 + l2)w − p. 4. Type 2.

Section 11. 1. Let G ⊂ Mob¨ (Cˆ ) be the group generated by isometries, dilatations with center at the origin, and one rection in a circle. Since G contains all translations, G also contains all dilatations with arbitrary center, and all rections in circles concentric to the given circle. Finally, using the translations again, G contains rections in all circles with arbitrary center. Thus, G Mob¨ (Cˆ ).

Section 12. 5. A linear fractional transformation that maps the unit disk D2 to H2 z → i i+z z →−i i−z the upper half-plane is i−z . Its inverse is i+z . 6. This transformation is a glide with axis the real axis, and the translation vector is 1. 7. Setting z |z|z(θ), the image of a circle |z| r<1 is an ellipse with semimajor axis r + 1/r and semiminor axis r − 1/r. The image of a half-line emanating from the origin is a hyperbola. The ellipses and the hyperbolas are confocal.

Section 13. 1. Problem 4 of Section 7 implies that the given linear fractional transformations are self maps of the unit disk D2. Theorem 9 asserts that the group of linear fractional transformations of H2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

434 Solutions for 100 Selected Problems

onto itself is 3-dimensional (with a, b, c, d ∈ R being the parameters subject to the constraint ad − bc 1). Since D2 and H2 are equivalent through a linear fractional transformation (Prob- lem 5 of Section 12), it follows that the group of linear fractional transformations that leave D2 invariant is also 3-dimensional. On the other hand, the linear fractional transformations given in the problem form a group, and they depend on 3 parameters, !(w), "(w), and θ. Thus, these give all the linear fractional transformations preserving D2. 4. We may assume that the triangle has vertices i, ti, t>1, and z(θ) <θ< π α β z(θ) ti ,0 2 . Let and be the angles at and at . (a) From the hyperbolic distance formula, we obtain t2 + 1 1 t2 + 1 cosh a , cosh b , cosh c . 2t sin θ 2t sin θ The Pythagorean theorem follows. (b) Let r>0 be the radius and c ∈ R the center of the semicircle that represents the hyperbolic line through ti and z(θ). Let δ be the angle at c between the radial segment connecting c and z(θ), and the real axis. We have t β ,rδ r β + θ, r δ θ. sin r cos cos cos sin sin The identity cos2 δ + sin2 δ 1 gives

r2 r2 cos2 β + 2r cos β cos θ + 1,

so that we have

t2 r2 sin2 β 2r cos β cos θ + 1.

Using this and α θ − δ, we compute 1 t t2 − 1 1 cos α cos(θ − δ) cos θ cos β + + r r 2t t sin β cosh a. Switching the roles of a, b and α, β, we also have

cos β sin α cosh b. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Solutions for 100 Selected Problems 435

Now the identity sinh a tan β tanh b follows by eliminating α.

Section 14. 1. (a) A parabolic isometry g is the composition of two rections in hyperbolic lines that meet at a common endpoint on the boundary of H2. Conjugating g with an isometry that carries this endpoint to ∞, the two hyperbolic lines become vertical Euclidean lines, and the conjugated g has the form z → z + a, a ∈ z → 1 z z → z + R. Now conjugate this with a to obtain 1. 2. Parabolic isometries have a unique fixed point on R ∪{∞}. El- liptic isometries have a unique fixed point in H2. Hyperbolic isometries have two fixed points on R ∪{∞}. 3. Since a linear fractional transformation determines the corresponding matrix in SL(2, C) up to sign, trace2 is well-defined. Given an isometry g, the condition az + b g(z) z, a, b, c, d ∈ , cz + d R for z to be a fixed point of g amounts to solving a quadratic equation with discriminant (a − d)2 − 4bc (a + d)2 − 4 trace2(g)− 4. If trace2(g) 4, then g has a unique fixed point on R ∪{∞}. Conjugating g by a suitable isometry, we may assume that this fixed point is ∞. This means that c 0 and b 0 (by unicity of the fixed point). As in Problem 1, we obtain that g is parabolic. If 0 ≤ trace2(g) < 4, then there are two fixed points of g, and they are conjugate complex numbers. Thus, there is a unique fixed point in H2. The argument in the text shows that g is elliptic. Finally, if trace2(g) > 4, then g has two fixed points on R ∪{∞}, which may be assumed to be 0 and ∞. We thus have b c 0, and g is hyperbolic. 5. Any positive integral power of a parabolic or hyperbolic isometry is of the same type. 6. If z0 ∈ R ∪{∞}is the unique fixed point of the parabolic g1, then by the stated commutativity, g2(z0) is left fixed by g1, and we must have g2(z0) z0. Thus, g2 is either parabolic or hyperbolic. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

436 Solutions for 100 Selected Problems

If w0 ∈ R∪{∞}were another fixed point of g2, then g1(w0) would also be left fixed by g2. This is a contradiction, since g1(w0) w0,z0. 7. As shown in the text, the conformal group corresponding to T T1 is SL(2, Z). It remains to consider the conformal group defined by T2. The composition of rections to the vertical sides of T2 gives z → z + 1. The composition of z →−¯z + 1 and z → 1/z¯ gives z →−1/z + 1, so that z →−1/z is also contained in this conformal group. 8. This follows from Example 8 by minor modifications.

Section 15. 2. Using the Euler formula for complex exponents, we have eiθ − e−iθ sin θ . 2i The complex extension of sine is therefore deﬁned by eiz − e−iz sin z . 2i Since the exponential function is periodic with period 2πi, the sine function is periodic with period 2π. To show that sin z is one-to-one on any vertical strip (k−1/2)π < !(z)<(k+1/2)π, k ∈ Z, write the difference of two sines as the product of a sine and a cosine, relate these to various complex exponentials, and recall the mapping properties of ez on horizontal strips.

Section 16.

2 1. (b) Let m0 ∈ S be the midpoint of the shorter great circular 2 arc connecting q0 and S(q0).IfS (q0)/∈ C, then consider the spherical triangle %m0S(q0)S(m0) and get a contradiction to the triangle inequality.

Section 17. 1. Label the vertices of the tetrahedron 1, 2, 3, 4. The three products of disjoint transpositions such as (12)(34) correspond to the three half-turns around the midpoints of the three pairs Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Solutions for 100 Selected Problems 437

of opposite edges. In addition, we have four 60◦ rotations corresponding to products such as (12)(13) (123), and four 120◦ rotations corresponding to products such as (13)(12) (132). 2. (a) Label the vertices of the tetrahedron such that p corresponds to 1, e connects 1 and 2, and f has vertices 1, 2, 3. Then S1 corresponds to the permutation (243), and S2 corresponds to (132). The product is (13)(24), and it corresponds to a half-turn. (c) Let S be the half-turn with axis connecting the midpoints of e1 and e2. The midpoints of the edges complementary to e1 and e2 form a quadrangle that has equal sides and parallel opposite edges. Since e1 and e2 are perpendicular, the quadrangle also has perpendicular adjacent edges, so it is a square. The quarter-turn around the axis of S followed by rection in the plane of the square carries the two pieces into one another. 3. The symmetry groups of T and its reciprocal are the same. 4. The vertices of a pair of reciprocal tetrahedra are those of a cube. Take a good look at Color Plate 5a. 5. (a) Notice√ that the midpoint of an edge in a unit square has distance 5/2 from an opposite vertex. + ∼ 7. (c) Use the fact that Symm (I) A5 is simple. For the image, compose φ with the even–odd homomorphism S5 →{±1}. 8. By the construction of the dodecahedron in the text, every vertex of the dodecahedron is the vertex of at most two cubes. The 5 inscribed cubes have the total of 40 vertices. Since the dodecahedron has 20 vertices, it follows that every vertex of the dodecahedron is the vertex of exactly two cubes. These two cubes have a common diagonal. There are exactly 20/2 10 diagonals. The number of different pairs of cubes is also 10. Thus, the statement is true. 9. No symmetry carries the polar vertices to the equatorial vertices. 13. Since the vertex ﬁgures are regular, all the faces meeting at a vertex must have equal side lengths and face angles. It follows that all the faces are congruent (and regular). Reciprocally, all vertex ﬁgures must be congruent (and regular). Thus, all dihedral angles of adjacent faces must be equal. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

438 Solutions for 100 Selected Problems

15. (a) The sum is a vector left fixed by the entire symmetry group of the Platonic solid. Thus, it must be zero. (b) Any symmetry carries midpoints of edges to midpoints of edges. (c) Same as in (b). (d) The two Platonic solids are reciprocal. This follows by considering their Schli¨ symbols. 16. Notice that at each vertex two hexagons and one pentagon meet. 17. (a) In Figure 17.28, the triangles %p0p3p4 and %p1p3p4 are similar (by the definition of the golden section). Thus all three angles at p3 are congruent and thereby equal to π/5. It follows that the segment connecting p0 and p4 is the side of a decagon inscribed in a unit circle. (b) We may assume that the line segment from the midpoint of a side (of the inscribed triangle) to the circle has unit length. Insert two crucial right triangles and study their intersection. 18. The vertices of the octahedron are the midpoints of the faces of the cube circumscribed around the reciprocal pair of tetrahedra. The vertex figure at a vertex v of the octahedron is parallel to the face of the cube whose midpoint is v. Homothety of the two cubes follows. Since the edges of the reciprocal cube bisect those of the octahedron perpendicularly, the ratio of 1 magnification is 2 . 19. The 20 vertices of the 5 tetrahedra are vertices of a regular polyhedron with symmetry group A5. Thus, it must be a dodecahedron. At a fixed vertex v of the colored icosahedron all 5 colors are represented. Consider the 5 triangular faces, one for each tetrahedron, that are extensions of the 5 icosahedral faces meeting at v. The 5 vertices closest to v in each of these triangular faces form a regular pentagon, since the order-5 rotation with axis through v permutes these vertices cyclically. The pentagon is perpendicular to this axis and hence parallel to the vertex figure of the icosahedron at v. Homothety of the two dodecahedra follows. 20. Take slices of the roof perpendicular to the base and use the Pythagorean theorem twice to conclude that the height of the roof is 1 − (τ/2)2 − ((τ − 1)/2)2 1/2. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Solutions for 100 Selected Problems 439

Look for similar right triangles and write the deﬁning equality of the golden section as τ/ / 2 1 2 . 1/2 (τ − 1)/2

3 22. Inscribe a Euclidean dodecahedron in D with vertices v1,..., 2 3 v20 ∈ S . By the hyperbolic geometry of D , for 0 π tan 0 2 1 2, so that 0 arctan 2 2 .By t ∈ ( , ) δ(t ) π continuity, there is a value 0 0 1 for which 0 2 . The fact that hyperbolic dodecahedra with this dihedral angle tesselate D3 follows, since at each vertex the three faces are mutually prependicular.

Section 19.

1. RP2#RP2 is homeomorphic to K2. Cut a hole in T2#RP2 so that it becomes the connected sum of a Mobius¨ strip and a torus. Cut a hole in RP2#RP2#RP2 K2#RP2 so that it becomes the connected sum of a Mobius¨ strip and a Klein bottle. Compare.

Section 20.

6. Use the Jordan curve theorem. 11. Find a state surrounded by an odd number of states. 12. If we assume that no more than three countries touch at one point (so that the corresponding graph is a triangulation), then a necessary and sufﬁcient condition is that each country have an even number of neighbors. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

440 Solutions for 100 Selected Problems

13. Four colors are needed for a checkerboard, but no finite number will suffice for every map. 14. Six. 15. Take a complete graph on 27 vertices, and color each edge red or green. Then we can find a subgraph that is either (i) a triangle with all edges colored green, or (ii) a complete graph on eight vertices with all edges colored red. 16. Use a stereographic projection to go between maps on the sphere and maps on the plane. One point (the point at infinity) will be missing from the sphere, but that does not affect the number of colors, because countries that touch at only one point are not required to have different colors. 17. Establish customs stations at every frontier between neighbor- ing countries, and build roads from each customs station to the capitals of both countries. If two roads should cross, re- build them by changing the connections so that they do not cross. The capitals are the vertices and the roads are the edges of a planar graph. (The roads should not cross any frontiers except at the customs stations. This can be done if every country is pathwise connected.) 18. The chromatic number is n. 19. Consider the graph whose vertices are those of a regular pentagon plus the centroid, and whose edges are the sides of the pentagon plus the five radial segments from the centroid to the vertices of the pentagon. 20. There cannot be a vertex of degree 3; else we could remove that vertex, color the remaining graph with four colors, replace the vertex, and color it differently from its three neighbors. If there is a vertex of degree 4, and we use four colors to color every other vertex, then the four neighbors of the vertex of degree 4 must require four different colors, say red, green, yellow, and blue; otherwise, the missing color could be used to color the vertex of degree 4. We can now use the same argument as in the proof of the five color theorem, looking at red–green paths or yellow–blue paths, and swapping either red and green or yellow and blue for part of the graph. 21. The red–blue path might cross the red–green path in Figure 20.6. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Solutions for 100 Selected Problems 441

Section 21.

2 1. Let v ∈ R , v 0. Since both J1v and J2v are perpendicular to v, J v ±J v J2v J 2v −v we have 1 2 . Moreover, 1 2 . Since a linear map is uniquely determined by its values on a basis, we have J1 ±J2. 2. Deﬁne Pαβ for α β, α, β 1,...,n, using the relations in the text.

Section 22. 1. The map is a rection in the plane R × R · k in V. 5. This follows by adjusting the argument in the text to the quaternionic case. The action of S3 ⊂ H on H2 is given by g : (p, q) → (p · g−1,q· g−1), g ∈ S3. 6. (a) The parallel of latitude at r corresponds to rotations with ﬁxed angle θ 2 cos−1(r). (b) The group of spherical rotations and S3/{±1} are isomorphic. SU(2) S3 is a subgroup of SL(2, C), and Mob¨ (Cˆ ) SL(2, C)/{±1}.

Section 23. 2. A rotation generating the orbit of the 5 tetrahedra circumscribed around the colored icosahedron has angle 2π/5. Its axis lies in the coordinate plane orthogonal to the ﬁrst axis, and the slope of the axis is the golden section τ (see Figure 17.30). This is because at the common vertex (0, 1,τ)of the icosahedron and a golden rectangle all 5 colors are represented. This axis goes through 1 τ 0, √ , √ . τ2 + 1 τ2 + 1 Using the notation in Problem 6 of Section 22, we have θ 2π/5 and π τ a cos , 5 2 τ π 1 b √ sin , τ2 + 1 5 2 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

442 Solutions for 100 Selected Problems 1 π 1 c √ sin , τ2 + 1 5 2τ d 0. Hence, the pair of quaternions that corresponds to this rotation is j ± 1 τ + i + . 2 τ

Section 26. 1. A vertex ﬁgure of the 4-dimensional cube is a regular 3- dimensional polyhedron with four vertices, since exactly four edges meet at a vertex. It is thus a regular tetrahedron. The 3-dimensional hyperplane through the origin with normal vector (1, 1, 1, 1) gives an octahedral slice of the cube; the four 3-dimensional cubic faces that meet at (1, 1, 1, 1) intersect this hyperplane in a regular tetrahedron, while the other four cubic faces meeting at (−1, −1, −1, −1) further truncate this tetrahedron to yield the octahedron. For an analogy, consider Problem 1 in Section 9. 2. (a) Take a good look at Figure 26.2. The Schli¨ symbol {4, 3, 3} means that the 3-dimensional faces of the 4-dimensional cube are ordinary 3-dimensional cubes with Schli¨ symbol {4, 3}, and each edge is surrounded by 3 of these. (b) Let Vn, En, and Fn denote the number of vertices, edges, and faces of the n- n dimensional cube. We have Vn 2 , En+1 2En + Vn, E1 1, and Fn+1 2Fn + En, F1 0. These recurrences can be solved n−1 n(n−1) n−2 En n Fn easily, and they give 2 and 2 2 . 3. (a) A 2-dimensional projection of the 4-dimensional tetrahedron is the pentagram star (top of Figure 20.4). (b) Each edge is surrounded by 3 ordinary tetrahedral faces; thus the Schli¨ symbol is {3, 3, 3}. 4. Consider the general form of a quadratic polynomial in 3 variables, and impose harmonicity. Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

...... Index . . . . .

Abel, N.H., 80 Aryabhatta, 253 A-conﬁguration, 86 Atlas, 183 extension of, 86 Action isotropy of, at a point, 402 Babylonians, 4 of a group on a set, 402 tablets of, 28, 72 orbit of, 402 Bachet, 36 Algebraic curve, 33 duplication formula of, 42–43 Alhambra, 119 Baker, A., 60 Amenta, N., 115 Bhaskara, 27 Analytic function, 174 Bimbo’s lozenge, 210 Angle of parallelism, 146 Binet, J.P.M., 251 Angular defect, 149 Birch–Swinnerton–Dyer conjecture, 32 Antipodal map, 205 Bolyai, J., 141, 149 Antiprism Boy, W., 272 pentagonal, 214 Boy’s surface, 272 square, 214 Brahmagupta, 16 triangular, 214 Branched covering, 92 Apollonius, 7 Branch Apollonius problem, 154 number, 277 Appel, K., 285 point, 277 Archimedes, 7, 15, 216 Bring, E.S., 348 Archimedes’s spiral, 245 Bring–Jerrard form, 348, 378 Argand diagram, 65 Brioschi quintic, 378 Ars Magna, 73 Buckyball, 251

443 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

444 Index

Bump-function, 390 projective space, 314 root, 69 sine function, 192 Canonical basis, 165 structure, 302 Carbon-60, 251 torus, 180 Cardano, G., 73 unit, 63 Cardinality, 398 vector space, 303 Caroll, L., 24 Conformality, 136, 175 Cartesian Bridge, 7 Connected sum, 275 Cartesian product Coordinate chart, 183 of sets, 397 Countable, 398 of topological spaces, 404 Critical point, 34 Casus irreducibilis, 81 Crystallographic restriction, 116 Cauchy, A., 40, 70, 316 Cube, 219 Cauchy formula, 65 Cuboctahedron, 235 Cauchy–Riemann equations, 175 Curve Cayley, A., 256, 285 algebraic, 33 Cell, 383 birationally equivalent, 35 16-cell, 385 elliptic, 38 24-cell, 385 rational, 33 120-cell, 386 Cyclic group, 4, 400 600-cell, 387 Change of bases formula, 301 Character, 333 D’Alembert, J., 65 Clebsch, 378 Da Vinci, L., 105, 214 Chord-method, 40 Delian problem, 82 Chrome alum, 219 De Morgan, A., 285 Clifford Descartes, R., 7, 216 algebra, 304 Diagonal surface, 378 module, 304 Diffeomorphism, 407 torus, 313 Dihedron, 103, 258 Complex Diophantus, 28 conjugation, 85 Dirichlet, P.G.L., 51 differentiable function, 173, 185 Discriminant, 73, 78, 420 exponential, 176 Distance function multiplication, 64, 66 Euclidean number, 64 of the plane, 97 absolute value of, 66 of the space, 210 argument of, 66 hyperbolic, of the plane, 144 imaginary part of, 64 in a metric space, 403 purely imaginary, 64 Dodecahedron, 219 real part of, 64 existence of, 223 plane, 66 Donkin, W.F., 99 projective plane, 195 Dorrie,¨ H., 23 projective quadric, 354 Double point, 34–35 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Index 445 e,10 Form, 332 irrationality of, 11, 12 dihedral, 335 Elliptic curve, 38 icosahedral, 341 Elliptic geometry, 205 octahedral, 339, 359 Elliptic integral, 194 tetrahedral, 337 of the ﬁrst kind, 194 Formula Equation, Bachet’s, 36 Cauchy, 65 Euclid’s Elements, 28, 70, 82, 84, 98, four square, 316 139, 222 Four-dimensional Eudoxus, 15 cube, 381 Euler, L., 6, 24, 31, 51, 65, 177, 203, 279 regular polytope, 383 Euler tetrahedron, 382 formula, 24, 177 Frey, G., 51 theorem for convex polyhedra, 216, curve, 51 250, 295 FTA, see Fundamental theorem of Euler–Poincare´ characteristic, 267, 294 algebra of the Klein bottle, 269 Fundamental of the projective plane, 270 domain, 120, 159 of the sphere, 267 parallelogram, 19 of the torus, 268 problem, 332 rational function, 332 set, 120, 159 Faltings, G., 51 Fundamental theorem of algebra, 65 Fermat, P., 6, 7, 31, 71, 316 proof of, 90–93 curve, 48 Fermat’s Last Theorem, 48, 69 Fermat’s method of inﬁnite descent, Galois, E., 80 49, 53 Galois theory, 419 Ferrari, L., 77 Gauss, C.F., 8, 17, 65, 141, 316 Ferro, S., 73 Gaussian integer, 70 Ferro–Fontana–Cardano formula, 80 Gelfond, A., 60 Fibonacci, 250 Generating line, 365 Field, 419 Geometric construction, 81 of complex numbers, 64 Geometric series formula, 10 extension, 419 Geometry degree of, 419 hyperbolic, 141 Galois, 420 Mobius,¨ 124 of rational functions, 57 Girard, A., 6, 65, 149 of rational numbers, 2 Glide, 98 of real numbers, 7 Golden quadratic, 419 rectangle, 246 splitting, 420 section, 244 Fifth Postulate, 139 Graph, 281 Fiore, A., 73 chromatic number of, 286 Flag, 212 circuit of, 281 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

446 Index

Graph (cont.) permutation, 401 complete, 294 point, 112 connected, 281 simple, 402 cycle of, 281 special orthogonal, 111, 320 directed, 295 special unitary, 256 edge of, 281 symmetric, on n letters, 401 edge-vertex matrix of, 295 tetrahedral, 232, 259 Eulerian, 281 torsion, 400 Hamiltonian, 282 triangle, 166 path of, 281 Guthrie, F., 284 planar, 283 reducible, 292 spanning cycle of, 282 Haken, W., 285 trail, 281 Hamilton, W.R., 307 unavoidable set in, 292 Hardy, G.H., 71 vertex of, 281 Harmonice Mundi, 120 degree of, 281 Heath, T.L., 28 walk of, 281 Heawood, P., 285 closed, 281 Heesch, H., 292 spanning, 281 Heptahedron, 270 Greatest integer function, 12 Hermite, C., 56, 349 Gregory, D., 17 Hessian, 337 Group, 399 Hilbert, D., 56 abelian, 399 Homeomorphism, 404 binary icosahedral, 265, 325 Homogeneous coordinate, 205 binary octahedral, 265, 325 Homomorphism, 400 binary tetrahedral, 265, 323 Hurwitz, A., 299 complex special linear, 135 Hyperbolic conformal, 166 angle, 145 crystallographic, 110 area, 148 generators of, 117–118 circle, 150 cyclic, 4, 400 distance function, 144 dihedral, 103, 229, 258 geometry, 141 discrete, 107 disk model of, 155 Fuchsian, 159 half-plane model of, 141 elementary, 163 line, 141 Galois, 420 space, 157 icosahedral, 238, 259 Hypercube, 383 Mobius,¨ 265 Hypergeometric Kleinian, 159 differential equation, 348, 351, 413 modular, 20, 159 function, 348, 413 tesselation by, 164 Hyppocrates of Chios, 27 of integers, 2 of integers modulo n,2 Mobius,¨ 127 Icosahedron, 219 octahedral, 235, 259 equation of, 248 orthogonal, 111, 320 symmetry group of, 238 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Index 447

Icosahedron book, 345 Level curve, 36 Icosidodecahedron, 235 Lindemann, F., 56 Indicial Liouville, J., 65, 186 equation, 411 Liouville’s theorem, 186 root, 411 Lobachevsky, N.I., 141, 149 Invariant, absolute, 333, 344 Lucas, 250 Isometric circle, 133 Isometry Euclidean Machin, J., 24–25 direct, 99 Manifold of the plane, 98 icosahedral, 328, 394 of the space, 210 octahedral, 328, 394 opposite, 99 tetrahedral, 328, 394 of the hyperbolic plane, 151 Map, 397 elliptic, 151 bilinear, 298 hyperbolic, 153 normed, 298 parabolic, 151 differential of, 407 Isomorphism, 400 Hopf, 313 identity, 398 Jacobian of, 407 Jacobian, 338 smooth, 407 Jerrard, G.B., 348 Veronese, 392 Maurolycus, 222 McKay, B., 285 Kali, 115, 119 Median, 86 Kempe, A., 285 Mersenne, M., 6 Kepler, J., 120, 219, 250 Method Klein, F., 56 chord-and-tangent, 40, 46 Klein bottle, 202 of exhaustion, 15 in four dimensions, 389 of rational slopes, 33 Konigsberg¨ bridge problem, tangent, 42 279 Metric space, 403 Kronecker, L., 7 Mobius¨ band Kummer, E.E., 51, 70 ﬁnite, 201 Kuratowski, K., 283 inﬁnite, 200 in four dimensions, 388 Lagrange, R.L., 65, 74, 316 Monic polynomial, 72 Lagrange substitution, 74, 365 Mordell, L.J., 45 Lambert, J.H., 14 Multiplication Lame,´ G., 51 complex, 64–66 Lattice, 25 Egyptian, 22 one-dimensional, 3 Greek, 22 two-dimensional square, 17 quaternionic, 307 Legendre, A.M., 51, 196 Leibniz, W., 17 Lens space, 238 Nautilus shell, 246 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

448 Index

Nicomachus, 22 Archimedes’ approximation of, 15, Normalformsatz, 363 24 nth roots of unity, 69 Babylonian approximation of, 14–15 Number Egyptian approximation of, 14–15 algebraic, 55 irrationality of, 14 complex, 64 Pierce, B., 177 congruent, 31, 39, 47–48 Plato, 29, 213, 219 constructible, 81 Platonic solid, 213 Gregory, 17, 24, 60 full symmetry group of, 242 irrational, 10 Playfair Axiom, 139 k-gonal, 22 Plutarch, 22 Liouville, 59 Poincare,´ H., 40 octahedral, 253 Poincare´ extension, 157 pentagonal, 22–23 Polygon, 209 polygonal, 22 regular n-sided, 3, 67, 96 square, 22 symmetries of, 101 square pyramid, 253 Polygonal numbers, 22 tetrahedral, 253 Polyhedral equations, 346 transcendental, 55 Polyhedron existence of, 57 centroid of, 212 triangular, 22–23 compound, 222 convex, 210 edge of, 210 Octahedron, 219 face of, 210 symmetry group of, 235 regular, 212 One-to-one correspondence, 398 vertex of, 210 Orbit, 103 Polynomial, homogeneous, 332 principal, 332 Projective special, 331 geometry, 205 Order ideal points in, 206 of a group, 399 line, 205 of an element in a group, 400 plane, 45, 205 Ordinary point, 409 models of, 205 Orthogonal multiplication, 299 point, 205 space, 208 Ptolemy, 82 Pacioli, Fra Luca, 214, 246 Pythagoras, 7 Parallel Pythagorean right-sensed, 146 theorem, 27 Pattern hyperbolic, 157 frieze, 114 triple, 27, 29 wallpaper, 117 Pentatope, 383 Period lattice, 194 Quadratic formula, 63 Permutation, 401 Quaternion, 307 π,10 conjugate, 308 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

Index 449

pure part of, 308 Schlegel diagram, 216 real part of, 308 Schneider, T., 60 Schwarz–Christoffel formula, 197 Schwarzian, 349, 417 Radical, 422 Schwarz’s lemma, 186 Radiolaria, 219 Screw displacement, 211 Radon, J., 299 Series Ramanujan, R., 16 geometric, 10 Rational curve, 33 Gregory–Leibniz, 17 degree of, 34 Sharp, A., 24 Rational point, 26 Singular point, 410 Real division algebra, 299 regular, 410 Reciprocal Sodium sulfantimoniate, 219 of a point, 106 Sphere, 83 of a polyhedron, 221 three-dimensional, 309 Rection Clifford tori in, 313 Euclidean Steiner, J., 271 of the plane, 98 Stella octangula, 222 Mobius¨ Stereographic projection, 83 in a circle, 125 Størmer, C., 17, 24 Regular polytope, 383 Subgroup, 400 Resolvent index of, 401 canonical, 362 left-coset of, 401 cubic, 79 Sumerians, 4, 72 Isosahedral, 372 Surface, 199 quintic, 356 triangulation of, 267 Rhind papyrus, 15 Ribet, K., 51 Richmond, H.W., 82 Taniyama–Shimura conjecture, 51 Riemann–Hurwitz relation, 278, 343 Tartaglia (Fontana), N., 73 Riemann moduli Tessellation of annuli, 189 Euclidean, 19 of complex tori, 187 hyperbolic, 159 Riemann surface, 183 by hyperbolic octagons, 169 conformal equivalence of, 186 spherical (Platonic), 213, 259 of genus p, 190 Tetrahedron, 219 Roman surface, 271 symmetry group of, 232 Roof-proof, 252 Thue, A., 36 Rotation, Euclidean, of the plane, 98 Timaeus of Locri, 219 Rotatory rection, 210, 234 Topological space, 403 Rufﬁni, P., 80 compact, 405 Hausdorff, 403 Total branch number, 277 Saddle condition, 35 Transformation Santa Claus, 83 complex linear fractional, 132 Schli¨ symbol, 215 conformal, 136, 175 Springer-Verlag Electronic Production toth 12:27 p.m. 2 · v · 2002

450 Index

Transformation (cont.) Van der Waerden, B.L., vii Mobius,¨ 127 Vertex ﬁgure, 221 side-pairing, 180 Viete,` F., 72, 80 Tschirnhaus, 74, 351 Translation, Euclidean, of the plane, Wantzel, P., 81 98 Weierstass form, 35, 194 Transposition, 401 Weierstass P-function, 194 Trisection, of an angle, 82 Wessell, C., 65 Tunnell, J., 31, 47 Wiles, A., 49 Wilson’s criterion, 70 Winding number, 94 Ultraradical, 348 Uniformization, 344 Unimodular, 20 Zagier, D., 32 Unit circle, 26, 67 Zeno’s paradox, 9